Chapter 1 · Some Basic Concepts of Chemistry

1.3 Stoichiometry: Formulae, Limiting Reagents and Concentration

From percentage composition to empirical and molecular formulae, from balanced equations to the masses that react, the reagent that runs out first, and four ways of saying how concentrated a solution is.

Every year acute diarrhoea kills hundreds of thousands of young children, mostly through dehydration. One of the cheapest life-savers in medicine is oral rehydration solution (ORS): a precise mixture of salt, glucose and potassium chloride in clean water. Too little salt and the gut cannot absorb the water; too much and the solution draws water out of the body. The recipe is written as amounts per litre, and getting it right is pure stoichiometry — the arithmetic of how much of each substance is present or reacts.

In Section 1.2 we learned to turn grams into moles. This section puts the mole to work. We find the formula of a compound from its percentage composition, read a balanced equation as a recipe in moles, predict the masses of reactants and products, find which reactant runs out first (the limiting reagent), and describe the composition of solutions by mass percent, mole fraction, molarity and molality. Throughout, subscript $1$ refers to the solvent and subscript $2$ to the solute.

Percentage Composition, Empirical and Molecular Formulae

When a chemist makes a new compound, one of the first measurements is its elemental analysis — the mass percentage of each element. Working backwards from these percentages gives the formula.

📘 Definition 1.10 — Mass percentage of an element

The mass percentage of an element in a compound is $$\text{mass }\%\text{ of element}=\frac{\text{(number of atoms of the element in the formula)}\times(\text{its atomic mass})}{\text{molar mass of the compound}}\times100.$$ The percentages of all the elements add up to $100\%$ (within rounding). The list of these percentages is the compound's percentage composition.

✏️ Example 1 — Percentage composition of ethanol

Find the mass percentage of carbon, hydrogen and oxygen in ethanol, $\mathrm{C_{2}H_{5}OH}$.

💡 Show solution▶

Count $\mathrm{C_{2}H_{5}OH}$ contains 2 C, 6 H and 1 O. Molar mass $=2(12.01)+6(1.008)+16.00=24.02+6.048+16.00=46.07\ \mathrm{g\,mol^{-1}}$.

Percentages C: $\dfrac{24.02}{46.07}\times100=52.14\%$; H: $\dfrac{6.048}{46.07}\times100=13.13\%$; O: $\dfrac{16.00}{46.07}\times100=34.73\%$.

Check $52.14+13.13+34.73=100.00\%$ ✓.

📘 Definition 1.11 — Empirical and molecular formula

The empirical formula of a compound gives the simplest whole-number ratio of the atoms of each element present. The molecular formula gives the exact number of atoms of each element in one molecule. For hydrogen peroxide the empirical formula is $\mathrm{HO}$ and the molecular formula $\mathrm{H_{2}O_{2}}$; for glucose they are $\mathrm{CH_{2}O}$ and $\mathrm{C_{6}H_{12}O_{6}}$; for water both are $\mathrm{H_{2}O}$. Percentage composition alone gives only the empirical formula — compounds with the same empirical formula (e.g. $\mathrm{C_{2}H_{2}}$ and $\mathrm{C_{6}H_{6}}$, both CH) have identical percentage compositions.

📐 Rule 1.12 — From empirical formula to molecular formula

The molecular formula is a whole-number multiple of the empirical formula: $$\text{molecular formula}=(\text{empirical formula})_{n},\qquad n=\frac{\text{molar mass of the compound}}{\text{empirical formula mass}},$$ where $n=1,2,3,\ldots$ The molar mass must be measured separately (for example by mass spectrometry).

🧭 Problem-solving strategy — empirical formula from percentages

Assume a $100\ \mathrm{g}$ sample, so each percentage becomes a mass in grams. (If an element's percentage is not given, it is $100$ minus the sum of the others — often oxygen.)
Convert each mass to moles of atoms: divide by the atomic mass.
Divide every mole value by the smallest of them.
If the ratios are not all close to whole numbers, multiply them all by the smallest integer that makes them so: a ratio ending in about $.5$ needs $\times2$, about $.33$ or $.67$ needs $\times3$, about $.25$ or $.75$ needs $\times4$. Round only ratios within about $0.1$ of an integer.
Write the empirical formula; if the molar mass is given, apply Rule 1.12.

✏️ Example 2 — Empirical and molecular formulae

(a) An oxide of iron contains $69.94\%$ iron. Find its empirical formula. (b) A sweet-smelling liquid contains $54.53\%$ C, $9.15\%$ H and the rest oxygen; its molar mass is $88.1\ \mathrm{g\,mol^{-1}}$. Find its empirical and molecular formulae.

💡 Show solution▶

(a) Moles In $100\ \mathrm{g}$: Fe $\dfrac{69.94}{55.85}=1.252\ \mathrm{mol}$; O $=100-69.94=30.06\ \mathrm{g}$, $\dfrac{30.06}{16.00}=1.879\ \mathrm{mol}$.

(a) Ratio Divide by $1.252$: $\mathrm{Fe:O}=1:1.500$. The $.5$ means multiply by 2: $2:3$. Empirical formula $\mathrm{Fe_{2}O_{3}}$. (Rounding $1.5$ to $2$ would give the wrong formula FeO$_{2}$!)

(b) Moles O $=100-54.53-9.15=36.32\%$. C $\dfrac{54.53}{12.01}=4.540$; H $\dfrac{9.15}{1.008}=9.077$; O $\dfrac{36.32}{16.00}=2.270\ \mathrm{mol}$.

(b) Ratio Divide by $2.270$: $\mathrm{C:H:O}=2.000:3.999:1.000$, so the empirical formula is $\mathrm{C_{2}H_{4}O}$ (mass $2(12.01)+4(1.008)+16.00=44.05\ \mathrm{g\,mol^{-1}}$).

(b) Molecular $n=\dfrac{88.1}{44.05}=2.00$, so the molecular formula is $\mathrm{(C_{2}H_{4}O)_{2}}=\mathrm{C_{4}H_{8}O_{2}}$ (ethyl ethanoate, a solvent in nail-polish remover, has this formula).

Balanced Equations and Stoichiometry

A balanced chemical equation is a statement about numbers of particles. Because a mole is just a fixed number of particles, the same coefficients also give the ratio of moles — but not of masses.

📘 Definition 1.12 — Stoichiometry

Stoichiometry (Greek stoicheion, element, and metron, measure) is the calculation of the quantities of reactants and products in a chemical reaction. The numbers in front of the formulae in a balanced equation are the stoichiometric coefficients. For $$\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(g)}$$ they say: $1$ molecule of methane reacts with $2$ molecules of oxygen to give $1$ molecule of carbon dioxide and $2$ molecules of water; equally, $1\ \mathrm{mol}$ $+$ $2\ \mathrm{mol}$ $\rightarrow$ $1\ \mathrm{mol}$ $+$ $2\ \mathrm{mol}$; and, in grams, $16.04\ \mathrm{g}+64.00\ \mathrm{g}\rightarrow44.01\ \mathrm{g}+36.03\ \mathrm{g}$ (both sides $80.04\ \mathrm{g}$ — conservation of mass). Each pair of coefficients gives a mole ratio, such as $\dfrac{2\ \mathrm{mol\ O_{2}}}{1\ \mathrm{mol\ CH_{4}}}$.

🧭 Problem-solving strategy — the stoichiometry road

Write the balanced equation, with state symbols. Check atoms of every element (and charge, if ions appear).
Convert the given quantity to moles ($n=m/M$, or $n=MV$ for a solution — see below).
Use the mole ratio from the coefficients to find moles of the wanted substance.
Convert to the unit asked for (mass, number of particles, volume of solution). Check that total mass is conserved.

$$\text{grams of A}\xrightarrow{\ \div M_{\mathrm{A}}\ }\text{mol A}\xrightarrow{\ \times\frac{\nu_{\mathrm{B}}}{\nu_{\mathrm{A}}}\ }\text{mol B}\xrightarrow{\ \times M_{\mathrm{B}}\ }\text{grams of B}$$

✏️ Example 3 — Burning cooking gas

Propane, $\mathrm{C_{3}H_{8}}$, burns completely in oxygen to give carbon dioxide and liquid water. For $22.0\ \mathrm{g}$ of propane, find (a) the mass of $\mathrm{CO_{2}}$ formed, (b) the mass of $\mathrm{O_{2}}$ used, (c) the mass of water formed, and check that mass is conserved.

💡 Show solution▶

Equation Balance C first, then H, then O: $$\mathrm{C_{3}H_{8}(g)+5O_{2}(g)\rightarrow3CO_{2}(g)+4H_{2}O(l)}$$ (C: 3 = 3; H: 8 = 8; O: 10 = 6 + 4 ✓).

Moles $M(\mathrm{C_{3}H_{8}})=3(12.01)+8(1.008)=44.09\ \mathrm{g\,mol^{-1}}$, so $n=\dfrac{22.0}{44.09}=0.4989\ \mathrm{mol}$.

(a) $n(\mathrm{CO_{2}})=3\times0.4989=1.497\ \mathrm{mol}$; $m=1.497\times44.01=65.9\ \mathrm{g}$.

(b) $n(\mathrm{O_{2}})=5\times0.4989=2.495\ \mathrm{mol}$; $m=2.495\times32.00=79.8\ \mathrm{g}$.

Check Reactants $22.0+79.8=101.8\ \mathrm{g}$; products $65.9+36.0=101.9\ \mathrm{g}$ — equal within rounding (with unrounded values both are $101.83\ \mathrm{g}$). ✓

The Limiting Reagent and Percentage Yield

In a sandwich shop with 10 slices of bread and 7 slices of cheese, at two slices of bread per sandwich only 5 sandwiches can be made — the bread runs out first, and 2 slices of cheese are left over. Reactions behave the same way: reactants are seldom mixed in exactly the ratio of the equation, and the one that runs out first decides how much product forms.

📘 Definition 1.13 — Limiting reagent, excess reagent and yield

The limiting reagent is the reactant that is completely consumed first; it limits the amount of product formed. The other reactants are excess reagents, and some of each is left over. The amount of product calculated from the limiting reagent is the theoretical yield; the amount actually obtained is the actual yield, and $$\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times100 .$$ Actual yields are usually below $100\%$ because of side reactions, incomplete reaction and losses during purification.

📐 Rule 1.13 — Finding the limiting reagent

For a reaction $a\,\mathrm{A}+b\,\mathrm{B}\rightarrow\text{products}$, convert each reactant to moles and divide by its coefficient. The reactant with the smaller value of $$\frac{n_{\mathrm{A}}}{a}\quad\text{or}\quad\frac{n_{\mathrm{B}}}{b}$$ is the limiting reagent, and that smaller value is the number of moles of "reaction sets" that can occur. If the two values are equal, the reactants are in the exact stoichiometric ratio and neither is in excess. Never compare the masses of the reactants, and never compare moles without dividing by the coefficients.

✏️ Example 4 — Aluminium chloride

$5.40\ \mathrm{g}$ of aluminium is heated with $14.2\ \mathrm{g}$ of chlorine: $\mathrm{2Al(s)+3Cl_{2}(g)\rightarrow2AlCl_{3}(s)}$. (a) Which is the limiting reagent? (b) What mass of aluminium chloride can form, and what mass of the excess reagent is left? (c) If $15.6\ \mathrm{g}$ of $\mathrm{AlCl_{3}}$ is actually collected, what is the percentage yield?

💡 Show solution▶

Moles $n(\mathrm{Al})=\dfrac{5.40}{26.98}=0.2001\ \mathrm{mol}$; $n(\mathrm{Cl_{2}})=\dfrac{14.2}{70.90}=0.2003\ \mathrm{mol}$. Almost equal moles — but the equation does not want them equal.

(a) Divide by the coefficients: Al $\dfrac{0.2001}{2}=0.1001$; $\mathrm{Cl_{2}}$ $\dfrac{0.2003}{3}=0.06676$. The smaller is chlorine: $\mathrm{Cl_{2}}$ is the limiting reagent.

(b) Product $n(\mathrm{AlCl_{3}})=\tfrac{2}{3}\times0.2003=0.1335\ \mathrm{mol}$; $M(\mathrm{AlCl_{3}})=26.98+3(35.45)=133.33\ \mathrm{g\,mol^{-1}}$; theoretical yield $=0.1335\times133.33=17.8\ \mathrm{g}$.

(b) Excess Al used $=\tfrac{2}{3}\times0.2003=0.1335\ \mathrm{mol}=3.60\ \mathrm{g}$, so $5.40-3.60=1.80\ \mathrm{g}$ of aluminium is left. Check: $14.2+3.60=17.8\ \mathrm{g}$ of product ✓.

Choose a reaction and how many particles of each reactant go into the flask (each particle stands for $0.1\ \mathrm{mol}$). Press ▶ Play: one "reaction set" at a time, the particles in the left flask are used up exactly as the balanced equation demands and the products appear on the right. The reaction stops when one reactant runs out — that is the limiting reagent; whatever is left (ringed) is in excess. Compare $n/\text{coefficient}$ in the readouts.

Reaction:

particles of A:10particles of B:7

A: $n$ and $n/\text{coeff.}$—

B: $n$ and $n/\text{coeff.}$—

limiting reagent—

product formed (mol)—

excess left (mol)—

⚠️ Common mistakes in stoichiometry

1. Using an unbalanced equation. Every mole ratio comes from the coefficients; one wrong coefficient spoils every number after it.

2. Using mass ratios as if they were mole ratios. In $\mathrm{2H_{2}+O_{2}\rightarrow2H_{2}O}$ the moles are $2:1$ but the masses are $4.03:32.00$, about $1:8$.

3. Choosing the limiting reagent by the smaller mass, or the smaller number of moles. In Example 4 chlorine is limiting although it is the larger mass and has slightly more moles than aluminium. Divide each amount by its coefficient (Rule 1.13).

4. Calculating the product from the excess reagent. The excess reagent is, by definition, not all used; only the limiting reagent tells you how much product forms.

Reactions in Solution: Concentration Terms

Most laboratory reactions — titrations, precipitations, the reactions in your body — take place in solution. A solution is a homogeneous mixture of a solute (the component present in the smaller amount, subscript $2$) dissolved in a solvent (the larger component, subscript $1$; usually water). Its concentration can be stated in several ways, each suited to a different job.

📘 Definition 1.14 — Ways of expressing concentration

For a solution containing $n_{2}$ moles (mass $w_{2}$) of solute in $n_{1}$ moles (mass $w_{1}$) of solvent, with total volume $V$:

Term	Definition	Unit
Mass percent (w/w)	$\dfrac{w_{2}}{w_{1}+w_{2}}\times100$	% (none)
Mole fraction of solute	$x_{2}=\dfrac{n_{2}}{n_{1}+n_{2}}$, with $x_{1}+x_{2}=1$	none
Molarity	$M=\dfrac{n_{2}}{V\ \text{of solution in L}}$	$\mathrm{mol\,L^{-1}}$ (written M)
Molality	$m=\dfrac{n_{2}}{\text{mass of solvent in kg}}$	$\mathrm{mol\,kg^{-1}}$ (written m)

Molarity is the most convenient in the laboratory (measure a volume and you know the moles: $n_{2}=M\times V$). Molality, mass percent and mole fraction depend only on masses, so they do not change with temperature; molarity does, because the volume of a solution expands slightly on warming.

⚠️ Molarity versus molality

Molarity divides by the volume of the whole solution; molality divides by the mass of the solvent only. A "$1\ \mathrm{M}$" solution is made by dissolving $1\ \mathrm{mol}$ of solute and adding water up to $1\ \mathrm{L}$ in a volumetric flask — not by adding the solute to $1\ \mathrm{L}$ of water. The symbols also clash: "M" means molarity, but $M_{2}$ (italic, with a subscript) is the molar mass of the solute; "m" means molality, not metres or mass. In dilute aqueous solutions (where $1\ \mathrm{L}$ of solution contains almost exactly $1\ \mathrm{kg}$ of water) molarity and molality are nearly equal; they part company as the solution becomes concentrated.

📐 Rule 1.14 — Dilution

Adding solvent changes the volume but not the amount of solute, so $n_{2}=M\times V$ is unchanged: $$M_{1}V_{1}=M_{2}V_{2},$$ where here (and only here) the subscripts $1$ and $2$ mean before and after dilution, not solvent and solute. Any volume unit may be used, provided both sides use the same one. The same reasoning — moles of solute simply add — gives the molarity of a mixture of two solutions of the same solute: $M_{\text{mix}}=\dfrac{M_{1}V_{1}+M_{2}V_{2}}{V_{1}+V_{2}}$ (assuming the volumes are additive).

✏️ Example 5 — Making and diluting a solution

(a) $4.00\ \mathrm{g}$ of sodium hydroxide is dissolved in water and the solution made up to $250.0\ \mathrm{mL}$ in a volumetric flask. Find its molarity. (b) What volume of this solution is needed to make $500.\ \mathrm{mL}$ of $0.0500\ \mathrm{M}$ NaOH?

💡 Show solution▶

(a) Moles $M(\mathrm{NaOH})=22.99+16.00+1.008=40.00\ \mathrm{g\,mol^{-1}}$; $n_{2}=\dfrac{4.00}{40.00}=0.100\ \mathrm{mol}$.

(a) Molarity $M=\dfrac{0.100\ \mathrm{mol}}{0.2500\ \mathrm{L}}=0.400\ \mathrm{mol\,L^{-1}}=0.400\ \mathrm{M}$.

(b) Dilution does not change the moles of solute (Rule 1.14): $M_{1}V_{1}=M_{2}V_{2}$ gives $$V_{1}=\frac{0.0500\ \mathrm{M}\times500.\ \mathrm{mL}}{0.400\ \mathrm{M}}=62.5\ \mathrm{mL}.$$ Pipette $62.5\ \mathrm{mL}$ of the stock solution into a $500\ \mathrm{mL}$ flask and make up to the mark with water.

📐 Rule 1.15 — Converting between concentration terms

Converting a volume-based quantity (molarity) into a mass-based one needs the density $d$ of the solution. Take $1\ \mathrm{L}$ of solution: its mass is $1000d$ grams ($d$ in $\mathrm{g\,mL^{-1}}$), of which $M\cdot M_{2}$ grams is solute, so

$$m=\frac{1000\,M}{1000\,d-M\,M_{2}}.$$

Converting molality to mole fraction needs no density: $1\ \mathrm{kg}$ of solvent is $\dfrac{1000}{M_{1}}$ mol, so $x_{2}=\dfrac{m}{m+1000/M_{1}}$ (for water, $1000/M_{1}=55.5\ \mathrm{mol}$).

✏️ Example 6 — One solution, four concentrations

$18.0\ \mathrm{g}$ of glucose ($M_{2}=180.16\ \mathrm{g\,mol^{-1}}$) is dissolved in $200.\ \mathrm{g}$ of water. The density of the solution is $1.03\ \mathrm{g\,mL^{-1}}$. Find (a) the mass percent of glucose, (b) the mole fraction of glucose, (c) the molality, (d) the molarity.

💡 Show solution▶

Moles $n_{2}=\dfrac{18.0}{180.16}=0.09991\ \mathrm{mol}$; $n_{1}=\dfrac{200.}{18.02}=11.10\ \mathrm{mol}$. Mass of solution $=218.0\ \mathrm{g}$.

(a) Mass percent $=\dfrac{18.0}{218.0}\times100=8.26\%$.

(b) $x_{2}=\dfrac{0.09991}{11.10+0.09991}=0.00892$ (and $x_{1}=0.991$) — even in a fairly sweet solution, fewer than one molecule in a hundred is glucose.

(d) Volume of solution $=\dfrac{218.0\ \mathrm{g}}{1.03\ \mathrm{g\,mL^{-1}}}=211.7\ \mathrm{mL}$; $M=\dfrac{0.09991}{0.2117\ \mathrm{L}}=0.472\ \mathrm{mol\,L^{-1}}$.

Check Rule 1.15: $m=\dfrac{1000(0.472)}{1000(1.03)-0.472(180.16)}=\dfrac{472}{945}=0.4996\approx0.500\ \mathrm{mol\,kg^{-1}}$ ✓. The molality is larger than the molarity because $1\ \mathrm{L}$ of this solution contains less than $1\ \mathrm{kg}$ of water.

Choose a solute in water and a mass percent. The graph shows the molality $m$ and the molarity $M$ of every solution from $0$ to $25\%$; the dots mark yours. The densities come from an approximate straight-line fit to measured values at $20\ ^{\circ}\mathrm{C}$. ▶ Play sweeps the mass percent. Notice that $m$ and $M$ agree for dilute solutions and separate as the solution gets more concentrated — and that for ethanol, which makes the solution less dense than water, the gap is largest.

Solute:mass %:10.0

in 100 g of solution—

density ($\mathrm{g\,mL^{-1}}$)—

mole fraction $x_{2}$—

molality $m$ ($\mathrm{mol\,kg^{-1}}$)—

molarity $M$ ($\mathrm{mol\,L^{-1}}$)—

Exercises

Use the atomic masses from Section 1.2 (H $1.008$, C $12.01$, N $14.01$, O $16.00$, Na $22.99$, Mg $24.31$, Al $26.98$, P $30.97$, S $32.06$, Cl $35.45$, K $39.10$, Ca $40.08$, Fe $55.85$, Zn $65.38$) unless told otherwise.

Percentage composition, empirical and molecular formulae.

1 Find the percentage composition of sulphuric acid, $\mathrm{H_{2}SO_{4}}$.
2 Which is the richer source of nitrogen as a fertiliser: urea, $\mathrm{CO(NH_{2})_{2}}$, or ammonium nitrate, $\mathrm{NH_{4}NO_{3}}$? Give the mass percentage of N in each.
3 What percentage of the mass of blue vitriol, $\mathrm{CuSO_{4}\cdot5H_{2}O}$ ($249.69\ \mathrm{g\,mol^{-1}}$), is water?
4 A compound contains $40.00\%$ C, $6.71\%$ H and $53.29\%$ O; its molar mass is $60.05\ \mathrm{g\,mol^{-1}}$. Find its empirical and molecular formulae.
5 A hydrocarbon contains $92.26\%$ carbon; its molar mass is $78.11\ \mathrm{g\,mol^{-1}}$. Find its molecular formula.
6 An oxide of phosphorus contains $43.64\%$ P. Its molar mass is $283.9\ \mathrm{g\,mol^{-1}}$. Find its empirical and molecular formulae.
7 An oxide of nitrogen contains $30.45\%$ nitrogen and has a molar mass of $92\ \mathrm{g\,mol^{-1}}$. Find its molecular formula.
8 $5.00\ \mathrm{g}$ of hydrated magnesium sulphate, $\mathrm{MgSO_{4}\cdot}x\mathrm{H_{2}O}$, is heated until all the water is driven off; $2.44\ \mathrm{g}$ of anhydrous $\mathrm{MgSO_{4}}$ remains. Find $x$.

Check: empirical-formula solver from percentage composition

element : mass %

molar mass (optional, g/mol):leave one % box blank for "the rest"

moles in 100 g—

÷ smallest—

empirical formula—

molecular formula—

💡 Show solutions▶

1 $M=98.08\ \mathrm{g\,mol^{-1}}$. H: $\dfrac{2.016}{98.08}\times100=2.06\%$; S: $\dfrac{32.06}{98.08}\times100=32.69\%$; O: $\dfrac{64.00}{98.08}\times100=65.26\%$.

2 Urea: $\dfrac{28.02}{60.06}\times100=46.65\%$ N; ammonium nitrate: $\dfrac{28.02}{80.04}\times100=35.01\%$ N. Urea is richer.

3 $\dfrac{5(18.016)}{249.69}\times100=\dfrac{90.08}{249.69}\times100=36.08\%$.

4 C $3.331$, H $6.657$, O $3.331\ \mathrm{mol}$; ratio $1:2.00:1$, empirical formula $\mathrm{CH_{2}O}$ ($30.03$). $n=\dfrac{60.05}{30.03}=2$: $\mathrm{C_{2}H_{4}O_{2}}$ (ethanoic acid).

5 H $=7.74\%$. C $\dfrac{92.26}{12.01}=7.682$, H $\dfrac{7.74}{1.008}=7.679$: ratio $1:1$, empirical formula CH ($13.02$). $n=\dfrac{78.11}{13.02}=6$: $\mathrm{C_{6}H_{6}}$ (benzene).

6 O $=56.36\%$. P $\dfrac{43.64}{30.97}=1.409$, O $\dfrac{56.36}{16.00}=3.523$: ratio $1:2.50$, so $\times2$: $\mathrm{P_{2}O_{5}}$ ($141.94$). $n=\dfrac{283.9}{141.94}=2$: $\mathrm{P_{4}O_{10}}$.

7 O $=69.55\%$. N $\dfrac{30.45}{14.01}=2.173$, O $\dfrac{69.55}{16.00}=4.347$: ratio $1:2.00$, empirical formula $\mathrm{NO_{2}}$ ($46.01$). $n=\dfrac{92}{46.01}=2$: $\mathrm{N_{2}O_{4}}$.

8 Water lost $=5.00-2.44=2.56\ \mathrm{g}=\dfrac{2.56}{18.02}=0.1421\ \mathrm{mol}$. $\mathrm{MgSO_{4}}$ ($120.37\ \mathrm{g\,mol^{-1}}$): $\dfrac{2.44}{120.37}=0.02027\ \mathrm{mol}$. $x=\dfrac{0.1421}{0.02027}=7.01\approx7$: $\mathrm{MgSO_{4}\cdot7H_{2}O}$ (Epsom salt).

Stoichiometry. Balance each equation first.

1 Potassium chlorate decomposes on heating: $\mathrm{KClO_{3}(s)\rightarrow KCl(s)+O_{2}(g)}$. What mass of oxygen is obtained from $12.25\ \mathrm{g}$ of $\mathrm{KClO_{3}}$?
2 What mass of quicklime, CaO, is produced by heating $1.00\ \mathrm{kg}$ of limestone, $\mathrm{CaCO_{3}}$?
3 How many moles of oxygen are needed to burn $2.0\ \mathrm{mol}$ of ethane completely: $\mathrm{C_{2}H_{6}(g)+O_{2}(g)\rightarrow CO_{2}(g)+H_{2}O(l)}$?
4 What mass of water forms when $8.0\ \mathrm{g}$ of methane burns completely?
5 What mass of zinc must react with excess hydrochloric acid to give $0.500\ \mathrm{g}$ of hydrogen? $\mathrm{Zn(s)+HCl(aq)\rightarrow ZnCl_{2}(aq)+H_{2}(g)}$
6 Industry. In a blast furnace, $\mathrm{Fe_{2}O_{3}(s)+CO(g)\rightarrow Fe(l)+CO_{2}(g)}$. What mass of iron can be obtained from $1.00$ tonne of $\mathrm{Fe_{2}O_{3}}$?
7 Biology. In photosynthesis $\mathrm{CO_{2}(g)+H_{2}O(l)\rightarrow C_{6}H_{12}O_{6}(aq)+O_{2}(g)}$. What mass of carbon dioxide is fixed when a plant makes $90.0\ \mathrm{g}$ of glucose?
8 $28.0\ \mathrm{g}$ of nitrogen reacts with excess hydrogen and $25.5\ \mathrm{g}$ of ammonia is obtained. Find the percentage yield.

💡 Show solutions▶

1 $\mathrm{2KClO_{3}(s)\rightarrow2KCl(s)+3O_{2}(g)}$. $n(\mathrm{KClO_{3}})=\dfrac{12.25}{122.55}=0.1000\ \mathrm{mol}$; $n(\mathrm{O_{2}})=\tfrac32\times0.1000=0.1500\ \mathrm{mol}$; $m=4.80\ \mathrm{g}$.

2 $\mathrm{CaCO_{3}(s)\xrightarrow{\Delta}CaO(s)+CO_{2}(g)}$, $1:1$. $\dfrac{1000}{100.09}=9.991\ \mathrm{mol}$; $m(\mathrm{CaO})=9.991\times56.08=560.\ \mathrm{g}$ ($5.60\times10^{2}\ \mathrm{g}$).

3 $\mathrm{2C_{2}H_{6}(g)+7O_{2}(g)\rightarrow4CO_{2}(g)+6H_{2}O(l)}$: $n(\mathrm{O_{2}})=\tfrac72\times2.0=7.0\ \mathrm{mol}$.

4 $\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)}$. $\dfrac{8.0}{16.04}=0.499\ \mathrm{mol}$; water $=2\times0.499\times18.02=18\ \mathrm{g}$ (two significant figures).

5 $\mathrm{Zn(s)+2HCl(aq)\rightarrow ZnCl_{2}(aq)+H_{2}(g)}$. $n(\mathrm{H_{2}})=\dfrac{0.500}{2.016}=0.2480\ \mathrm{mol}=n(\mathrm{Zn})$; $m=0.2480\times65.38=16.2\ \mathrm{g}$.

6 $\mathrm{Fe_{2}O_{3}(s)+3CO(g)\rightarrow2Fe(l)+3CO_{2}(g)}$. $\dfrac{1.00\times10^{6}\ \mathrm{g}}{159.70\ \mathrm{g\,mol^{-1}}}=6262\ \mathrm{mol}$; Fe $=2\times6262\times55.85=6.99\times10^{5}\ \mathrm{g}=699\ \mathrm{kg}$.

7 $\mathrm{6CO_{2}(g)+6H_{2}O(l)\rightarrow C_{6}H_{12}O_{6}(aq)+6O_{2}(g)}$. $\dfrac{90.0}{180.16}=0.4996\ \mathrm{mol}$ glucose; $\mathrm{CO_{2}}=6\times0.4996\times44.01=132\ \mathrm{g}$.

8 $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow2NH_{3}(g)}$. $\dfrac{28.0}{28.02}=0.9993\ \mathrm{mol}$ $\mathrm{N_{2}}$; theoretical $\mathrm{NH_{3}}=2\times0.9993\times17.03=34.04\ \mathrm{g}$; yield $=\dfrac{25.5}{34.04}\times100=74.9\%$.

III

Limiting reagent. In each part name the limiting reagent.

1 $4.0\ \mathrm{g}$ of hydrogen is exploded with $16.0\ \mathrm{g}$ of oxygen. What mass of water forms, and what mass of which gas remains?
2 $3.0\ \mathrm{mol}$ of $\mathrm{N_{2}}$ is mixed with $6.0\ \mathrm{mol}$ of $\mathrm{H_{2}}$. What is the maximum amount of ammonia, and how much of the excess reagent remains?
3 $10.0\ \mathrm{g}$ of $\mathrm{CaCO_{3}}$ is added to a solution containing $7.30\ \mathrm{g}$ of HCl: $\mathrm{CaCO_{3}(s)+2HCl(aq)\rightarrow CaCl_{2}(aq)+H_{2}O(l)+CO_{2}(g)}$. What mass of $\mathrm{CO_{2}}$ is given off?
4 $2.43\ \mathrm{g}$ of magnesium is added to a solution containing $0.150\ \mathrm{mol}$ of HCl. What mass of hydrogen forms, and what mass of magnesium is left?
5 A box contains 6 molecules of $\mathrm{N_{2}}$ and 12 molecules of $\mathrm{H_{2}}$. After the reaction to form ammonia is complete, how many molecules of each substance are present?
6 Engineering. In the thermite reaction used to weld railway tracks, $\mathrm{2Al(s)+Fe_{2}O_{3}(s)\rightarrow Al_{2}O_{3}(s)+2Fe(l)}$. A charge contains $54.0\ \mathrm{g}$ of Al and $200.\ \mathrm{g}$ of $\mathrm{Fe_{2}O_{3}}$. What mass of molten iron forms, and what mass of the excess reagent is left?
7 Challenge. Equal masses, $10.0\ \mathrm{g}$ each, of zinc and sulphur are heated together to form zinc sulphide, ZnS. What mass of ZnS forms, and what mass of which element is left over?

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1 $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow2H_{2}O(l)}$. $\mathrm{H_{2}}$: $\dfrac{4.0}{2.016}=1.98\ \mathrm{mol}$, $\div2=0.99$; $\mathrm{O_{2}}$: $0.500\ \mathrm{mol}$, $\div1=0.500$. Oxygen is limiting. Water $=1.000\ \mathrm{mol}=18.0\ \mathrm{g}$; $\mathrm{H_{2}}$ used $1.000\ \mathrm{mol}=2.016\ \mathrm{g}$, so $4.0-2.016=2.0\ \mathrm{g}$ of hydrogen remains.

2 $\mathrm{N_{2}}$: $3.0/1=3.0$; $\mathrm{H_{2}}$: $6.0/3=2.0$. Hydrogen is limiting. $\mathrm{NH_{3}}=\tfrac23\times6.0=4.0\ \mathrm{mol}$; $\mathrm{N_{2}}$ used $2.0\ \mathrm{mol}$, so $1.0\ \mathrm{mol}$ $\mathrm{N_{2}}$ remains.

3 $\mathrm{CaCO_{3}}$: $\dfrac{10.0}{100.09}=0.09991$, $\div1=0.09991$; HCl: $\dfrac{7.30}{36.46}=0.2002$, $\div2=0.1001$. $\mathrm{CaCO_{3}}$ is limiting — but only just. $\mathrm{CO_{2}}=0.09991\times44.01=4.40\ \mathrm{g}$ (HCl left: only $4\times10^{-4}\ \mathrm{mol}$, about $0.015\ \mathrm{g}$).

4 $\mathrm{Mg(s)+2HCl(aq)\rightarrow MgCl_{2}(aq)+H_{2}(g)}$. Mg: $\dfrac{2.43}{24.31}=0.09996$; HCl: $0.150/2=0.0750$. HCl is limiting. $\mathrm{H_{2}}=0.0750\ \mathrm{mol}=0.151\ \mathrm{g}$; Mg left $=(0.09996-0.0750)\times24.31=0.607\ \mathrm{g}$.

5 $\mathrm{N_{2}+3H_{2}\rightarrow2NH_{3}}$: 12 $\mathrm{H_{2}}$ allow $12/3=4$ sets; 6 $\mathrm{N_{2}}$ would allow 6. $\mathrm{H_{2}}$ is limiting. After: 8 $\mathrm{NH_{3}}$, $6-4=2$ $\mathrm{N_{2}}$, 0 $\mathrm{H_{2}}$. (Atoms: N $12=8+4$ ✓, H $24=24$ ✓.) Try it in Interactive 1.

6 Al: $\dfrac{54.0}{26.98}=2.001$, $\div2=1.001$; $\mathrm{Fe_{2}O_{3}}$: $\dfrac{200.}{159.70}=1.252$, $\div1=1.252$. Al is limiting. Fe $=2.001\times55.85=112\ \mathrm{g}$; $\mathrm{Fe_{2}O_{3}}$ used $1.001\times159.70=159.8\ \mathrm{g}$, left $40.2\ \mathrm{g}$.

7 $\mathrm{Zn(s)+S(s)\rightarrow ZnS(s)}$. Zn $\dfrac{10.0}{65.38}=0.1530\ \mathrm{mol}$; S $\dfrac{10.0}{32.06}=0.3119\ \mathrm{mol}$. Zinc is limiting. ZnS ($97.44\ \mathrm{g\,mol^{-1}}$) $=0.1530\times97.44=14.9\ \mathrm{g}$; sulphur left $=10.0-0.1530\times32.06=5.10\ \mathrm{g}$. Check: $14.9+5.10=20.0\ \mathrm{g}$ ✓.

Concentration of solutions (subscript 1 = solvent, 2 = solute).

1 Find the molarity of a solution made by dissolving $5.85\ \mathrm{g}$ of NaCl in water and making it up to $500.\ \mathrm{mL}$.
2 What mass of KOH is needed to make $250.\ \mathrm{mL}$ of $0.150\ \mathrm{M}$ solution?
3 What volume of $12.0\ \mathrm{M}$ hydrochloric acid is needed to make $1.00\ \mathrm{L}$ of $0.300\ \mathrm{M}$ acid?
4 Find the molality of a solution of $6.00\ \mathrm{g}$ of urea, $\mathrm{CO(NH_{2})_{2}}$, in $250.\ \mathrm{g}$ of water.
5 $46.0\ \mathrm{g}$ of ethanol is mixed with $36.0\ \mathrm{g}$ of water. Find the mole fraction of each.
6 $25.0\ \mathrm{g}$ of sugar is dissolved in $100.\ \mathrm{g}$ of water. Find the mass percent of sugar.
7 Concentrated sulphuric acid is $98.0\%$ $\mathrm{H_{2}SO_{4}}$ by mass and has density $1.84\ \mathrm{g\,mL^{-1}}$. Find its molarity and its molality.
8 A $2.00\ \mathrm{M}$ NaCl solution has density $1.08\ \mathrm{g\,mL^{-1}}$. Find its molality.
9 The mole fraction of urea in an aqueous solution is $0.0500$. Find the molality.
10 Medicine. A saline drip is $0.90\%$ (w/w) NaCl with density $1.005\ \mathrm{g\,mL^{-1}}$. Find its molarity and molality to two significant figures.
11 Explain why molality, not molarity, is used when a solution's properties are studied over a range of temperatures.

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1 $n_{2}=\dfrac{5.85}{58.44}=0.1001\ \mathrm{mol}$; $M=\dfrac{0.1001}{0.500}=0.200\ \mathrm{M}$.

2 $n_{2}=0.150\times0.250=0.0375\ \mathrm{mol}$; $M(\mathrm{KOH})=56.11\ \mathrm{g\,mol^{-1}}$; $m=2.10\ \mathrm{g}$.

3 $V_{1}=\dfrac{0.300\times1.00}{12.0}=0.0250\ \mathrm{L}=25.0\ \mathrm{mL}$ (add the acid to water, never water to the concentrated acid).

4 $n_{2}=\dfrac{6.00}{60.06}=0.09990\ \mathrm{mol}$; $m=\dfrac{0.09990}{0.250}=0.400\ \mathrm{mol\,kg^{-1}}$.

5 $n(\text{ethanol})=\dfrac{46.0}{46.07}=0.9985$; $n(\text{water})=\dfrac{36.0}{18.02}=1.998$. $x(\text{ethanol})=\dfrac{0.9985}{2.997}=0.333$; $x(\text{water})=0.667$.

6 $\dfrac{25.0}{125.0}\times100=20.0\%$.

7 $1\ \mathrm{L}$ has mass $1840\ \mathrm{g}$, of which $0.980\times1840=1803\ \mathrm{g}$ is $\mathrm{H_{2}SO_{4}}$: $\dfrac{1803}{98.08}=18.4\ \mathrm{mol}$, so $M=18.4\ \mathrm{M}$. Per $100\ \mathrm{g}$: $98.0\ \mathrm{g}$ acid $=0.9992\ \mathrm{mol}$ in $2.0\ \mathrm{g}=0.0020\ \mathrm{kg}$ water: $m\approx5.0\times10^{2}\ \mathrm{mol\,kg^{-1}}$ — molality becomes huge when there is almost no solvent.

8 $1\ \mathrm{L}$: mass $1080\ \mathrm{g}$, NaCl $2.00\times58.44=116.9\ \mathrm{g}$, water $963.1\ \mathrm{g}$. $m=\dfrac{2.00}{0.9631}=2.08\ \mathrm{mol\,kg^{-1}}$.

9 Take $1\ \mathrm{mol}$ in total: $0.0500\ \mathrm{mol}$ urea and $0.9500\ \mathrm{mol}$ water $=0.9500\times18.02=17.11\ \mathrm{g}$. $m=\dfrac{0.0500}{0.01711}=2.92\ \mathrm{mol\,kg^{-1}}$.

10 $1\ \mathrm{L}$ has mass $1005\ \mathrm{g}$ and contains $9.045\ \mathrm{g}$ NaCl $=0.1548\ \mathrm{mol}$: $M=0.15\ \mathrm{M}$. Per $1000\ \mathrm{g}$ of solution: $9.0\ \mathrm{g}$ NaCl $=0.1540\ \mathrm{mol}$ in $0.991\ \mathrm{kg}$ water: $m=0.1554\to0.16\ \mathrm{mol\,kg^{-1}}$.

11 Molality depends only on masses, which do not change with temperature; molarity uses the solution's volume, which expands on heating, so the same solution has a slightly different molarity at each temperature.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET A compound contains $50\%$ by mass of element X (atomic mass $10$) and $50\%$ of element Y (atomic mass $20$). Its empirical formula is (A) XY (B) $\mathrm{X_{2}Y}$ (C) $\mathrm{XY_{2}}$ (D) $\mathrm{X_{2}Y_{3}}$

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Moles In $100\ \mathrm{g}$: X $\dfrac{50}{10}=5$; Y $\dfrac{50}{20}=2.5$.

Answer Ratio $5:2.5=2:1$: (B) $\mathrm{X_{2}Y}$. Equal masses do not mean equal numbers of atoms.

E2. JEE Main Numerical answer. The mole fraction of solute in a $2.5\ \mathrm{mol\,kg^{-1}}$ aqueous solution is $y\times10^{-2}$. Find $y$ to the nearest integer. ($M_{1}=18.02\ \mathrm{g\,mol^{-1}}$)

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Take 1 kg water $n_{2}=2.5\ \mathrm{mol}$; $n_{1}=\dfrac{1000}{18.02}=55.5\ \mathrm{mol}$.

Mole fraction $x_{2}=\dfrac{2.5}{2.5+55.5}=0.0431=4.31\times10^{-2}$.

Answer $y=4$.

E3. Board · 3 marks $128\ \mathrm{g}$ of sulphur dioxide and $64.0\ \mathrm{g}$ of oxygen are heated with a catalyst: $\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow2SO_{3}(g)}$. (i) Identify the limiting reagent. (ii) Calculate the maximum mass of $\mathrm{SO_{3}}$. (iii) What mass of the excess reagent remains?

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(i) 1 mark $n(\mathrm{SO_{2}})=\dfrac{128}{64.06}=1.998\ \mathrm{mol}$, $\div2=0.999$; $n(\mathrm{O_{2}})=\dfrac{64.0}{32.00}=2.00\ \mathrm{mol}$, $\div1=2.00$. $\mathrm{SO_{2}}$ is limiting.

(ii) 1 mark $n(\mathrm{SO_{3}})=n(\mathrm{SO_{2}})=1.998\ \mathrm{mol}$; $m=1.998\times80.06=160.\ \mathrm{g}$.

(iii) 1 mark $\mathrm{O_{2}}$ used $=0.999\ \mathrm{mol}=32.0\ \mathrm{g}$; left $=64.0-32.0=32.0\ \mathrm{g}$ of oxygen.

E4. JEE Advanced Numerical answer. A $2.50\ \mathrm{M}$ aqueous solution of NaCl has a density of $1.10\ \mathrm{g\,mL^{-1}}$. Find its molality, in $\mathrm{mol\,kg^{-1}}$, to two decimal places.

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Take 1 L Mass of solution $=1100\ \mathrm{g}$; NaCl $=2.50\times58.44=146.1\ \mathrm{g}$; water $=1100-146.1=953.9\ \mathrm{g}$.

Molality $m=\dfrac{2.50\ \mathrm{mol}}{0.9539\ \mathrm{kg}}=2.62\ \mathrm{mol\,kg^{-1}}$. (Rule 1.15 gives the same: $\dfrac{1000(2.50)}{1100-2.50(58.44)}=2.62$.)

Answer $2.62$. The common error is dividing by the mass of solution ($1.100\ \mathrm{kg}$), giving $2.27$.

E5. NEET JEE Main What volume of water must be added to $200\ \mathrm{mL}$ of $0.50\ \mathrm{M}$ NaOH to make it $0.20\ \mathrm{M}$? (A) $500\ \mathrm{mL}$ (B) $300\ \mathrm{mL}$ (C) $200\ \mathrm{mL}$ (D) $80\ \mathrm{mL}$

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Dilution $V_{2}=\dfrac{M_{1}V_{1}}{M_{2}}=\dfrac{0.50\times200}{0.20}=500\ \mathrm{mL}$ — this is the final volume.

Answer Water to add $=500-200=300\ \mathrm{mL}$: (B). Option (A) is the trap.

E6. Board · Case study An antacid tablet contains $0.500\ \mathrm{g}$ of magnesium hydroxide, which neutralises stomach acid: $\mathrm{Mg(OH)_{2}(s)+2HCl(aq)\rightarrow MgCl_{2}(aq)+2H_{2}O(l)}$. (i) How many moles of HCl can one tablet neutralise? (2 marks) (ii) Taking stomach acid as $0.10\ \mathrm{M}$ HCl, what volume of it does one tablet neutralise? (1 mark) (iii) What mass of magnesium chloride forms? (1 mark)

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(i) $M(\mathrm{Mg(OH)_{2}})=24.31+2(16.00+1.008)=58.32\ \mathrm{g\,mol^{-1}}$; $n=\dfrac{0.500}{58.32}=8.573\times10^{-3}\ \mathrm{mol}$. Each mole neutralises 2 mol HCl: $n(\mathrm{HCl})=1.715\times10^{-2}\ \mathrm{mol}$.

(ii) $V=\dfrac{n}{M}=\dfrac{1.715\times10^{-2}\ \mathrm{mol}}{0.10\ \mathrm{mol\,L^{-1}}}=0.17\ \mathrm{L}$ (about $170\ \mathrm{mL}$; two significant figures, set by $0.10\ \mathrm{M}$).

(iii) $n(\mathrm{MgCl_{2}})=8.573\times10^{-3}\ \mathrm{mol}$; $M=95.21\ \mathrm{g\,mol^{-1}}$; $m=0.816\ \mathrm{g}$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why does the World Health Organization's "reduced-osmolarity" oral rehydration solution contain exactly the amounts it does — and what would a mixing mistake do to its concentrations?

Why it matters. Oral rehydration therapy is credited with saving tens of millions of lives since the 1970s, from Bangladesh and India to sub-Saharan Africa. In 2002 the WHO and UNICEF changed the recipe to a lower total concentration, which reduced vomiting and the need for intravenous drips. Health workers in many countries mix ORS from sachets — and a sachet dissolved in the wrong volume changes every concentration.

What to do.

One litre of the WHO solution contains $2.6\ \mathrm{g}$ NaCl, $13.5\ \mathrm{g}$ anhydrous glucose, $1.5\ \mathrm{g}$ KCl and $2.9\ \mathrm{g}$ trisodium citrate dihydrate, $\mathrm{Na_{3}C_{6}H_{5}O_{7}\cdot2H_{2}O}$. Calculate the molar mass of each and the molarity of each, in $\mathrm{mmol\,L^{-1}}$.
Each formula unit gives ions in solution: NaCl gives $\mathrm{Na^{+}}+\mathrm{Cl^{-}}$, KCl gives $\mathrm{K^{+}}+\mathrm{Cl^{-}}$, the citrate gives $3\mathrm{Na^{+}}$ and one citrate ion. Find the total concentrations of $\mathrm{Na^{+}}$, $\mathrm{K^{+}}$ and $\mathrm{Cl^{-}}$.
Add up the concentrations of all the dissolved particles (ions and glucose molecules). The WHO quotes about $245\ \mathrm{mmol\,L^{-1}}$. Do you agree?
A parent dissolves a one-litre sachet in a $750\ \mathrm{mL}$ bottle of water by mistake. Use Rule 1.14 to find the new total concentration, and compare it with blood plasma (about $290\ \mathrm{mmol\,L^{-1}}$ of dissolved particles).

Tools: Calculator or spreadsheet; Rule 1.11 and Definition 1.14.

Hand in: A table of molar masses and concentrations, your total, the result for the mixing mistake, and a two-sentence explanation of why the glucose and sodium concentrations are almost equal (find out about sodium–glucose co-transport in the gut).

Going further. Hospitals express saline drips as $0.9\%$ "w/v" (grams per $100\ \mathrm{mL}$ of solution). How does w/v differ from the w/w mass percent of this section, and how big is the difference for such a dilute solution?

Instructor note: molar masses NaCl $58.44$, glucose $180.16$, KCl $74.55$, citrate dihydrate $294.10\ \mathrm{g\,mol^{-1}}$; concentrations $44.5$, $74.9$, $20.1$, $9.86\ \mathrm{mmol\,L^{-1}}$. $\mathrm{Na^{+}}=44.5+3(9.86)=74.1$, $\mathrm{K^{+}}=20.1$, $\mathrm{Cl^{-}}=64.6$, citrate $9.9$, glucose $74.9$; total $\approx244\ \mathrm{mmol\,L^{-1}}$ (WHO rounds its targets to $75$, $20$, $65$, $10$, $75$, total $245$). In $0.750\ \mathrm{L}$ every concentration rises by $\tfrac{1}{0.750}$: total $\approx325\ \mathrm{mmol\,L^{-1}}$, above plasma — the solution would draw water into the gut instead of out of it.

🧪 Self-check quiz

0/5 correct

Q1. The empirical formula of glucose, $\mathrm{C_{6}H_{12}O_{6}}$, is

Q2. $2.0\ \mathrm{mol}$ of $\mathrm{H_{2}}$ is mixed with $2.0\ \mathrm{mol}$ of $\mathrm{O_{2}}$ and the mixture is exploded: $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow2H_{2}O(l)}$. Which statement is correct?

Q3. $0.50\ \mathrm{mol}$ of a solute is dissolved in water to make $250\ \mathrm{mL}$ of solution. Its molarity is

Q4. Which concentration term changes when a solution is warmed?

Q5. $100\ \mathrm{mL}$ of $1.0\ \mathrm{M}$ HCl is diluted to $0.25\ \mathrm{M}$. The final volume of the solution is

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