Empirical Formula Practice

Every question on this page is the same question. You are given masses or percentages, which is what a balance can tell you, and you are asked for a ratio of counts, which is what a formula is. The mole is the only bridge between those two things, and the route never changes:

1. Assume 100 g if you were given percentages, so percentages become grams.
2. Divide each mass by that element’s molar mass, to get moles.
3. Divide every result by the smallest of them.
4. If what comes out is not close to whole numbers, multiply all of them by a small integer until it is.

Molar masses to two decimal places. Keep full precision through steps 2 and 3 — rounding early is what turns a genuine 1.50 into a fake 1.5 you cannot interpret.

1. A compound is found to be 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

Answer 1

Take 100.0 g of the compound, so the percentages become grams directly. That assumption costs nothing, because a ratio does not care how much you started with.

$\begin{array}{rl}n(\mathrm{C})& =\frac{40.0}{12.01}=3.3306\text{ mol}\\ n(\mathrm{H})& =\frac{6.7}{1.01}=6.6337\text{ mol}\\ n(\mathrm{O})& =\frac{53.3}{16.00}=3.3313\text{ mol}\end{array}$

Divide each by the smallest, which is 3.3306:

$\begin{array}{rl}\mathrm{C}& :\frac{3.3306}{3.3306}=1.000\\ \mathrm{H}& :\frac{6.6337}{3.3306}=1.992\\ \mathrm{O}& :\frac{3.3313}{3.3306}=1.000\end{array}$

The empirical formula is $\mathrm{CH}{}_2\mathrm{O}$.

The hydrogen came out at 1.992 rather than exactly 2. That is a rounding of the input percentages, not a chemical fact — 6.7% is given to two significant figures, and two figures cannot produce a ratio good to four. A value within about a percent of a whole number is a whole number.

2. The compound in question 1 has a molar mass of 180.2 g/mol. Find its molecular formula.

Answer 2

The empirical formula gives the ratio; the molar mass gives the size. You need both, and neither one alone is enough.

$M(\mathrm{CH}{}_2\mathrm{O})=12.01+2(1.01)+16.00=30.03\text{ g/mol}$

$\frac{180.2}{30.03}=6.001$

So the molecule is six empirical units:

$\mathrm{C}{}_6\mathrm{H}{}_{12}\mathrm{O}{}_6$

That is glucose. Notice that the empirical formula $\mathrm{CH}{}_2\mathrm{O}$ is also the empirical formula of formaldehyde, of acetic acid, and of ribose — all of which have the same carbon, hydrogen, and oxygen ratio and none of which is remotely the same substance. The ratio was never going to be enough on its own, which is exactly the point of Empirical and Molecular Formulas.

3. A strip of magnesium of mass 0.486 g is heated in a crucible until it will react no further. The product has a mass of 0.806 g. Find the empirical formula of the product.

Answer 3

The oxygen was never weighed directly. It is the difference, and saying so out loud is the whole trick of this style of question.

$m(\mathrm{O})=0.806\text{ g}-0.486\text{ g}=0.320\text{ g}$

$\begin{array}{rl}n(\mathrm{Mg})& =\frac{0.486}{24.31}=0.019992\text{ mol}\\ n(\mathrm{O})& =\frac{0.320}{16.00}=0.020000\text{ mol}\end{array}$

Divide both by the smaller, 0.019992:

$\mathrm{Mg}:1.000\mathrm{O}:1.000$

The empirical formula is $\mathrm{MgO}$.

Two things worth noticing. The oxygen mass, 0.320 g, is a difference of two measured masses, so its uncertainty is larger than either of theirs — differences of similar numbers always lose precision. And the answer is only trustworthy because the magnesium was heated until the mass stopped changing; a strip taken off the flame early would leave unreacted magnesium in the crucible, the oxygen difference would come out too small, and the formula would come out wrong in a predictable direction.

4. A sample of hydrated copper(II) sulfate has a mass of 4.99 g. After heating to constant mass, 3.19 g of the anhydrous solid remains. Find $x$ in $\mathrm{CuSO}{}_4\cdot x\mathrm{H}{}_2\mathrm{O}$.

Answer 4

$M(\mathrm{CuSO}{}_4)=63.55+32.07+4(16.00)=159.62\text{ g/mol}$

$M(\mathrm{H}{}_2\mathrm{O})=18.02\text{ g/mol}$

The water lost is the difference in mass:

$m(\mathrm{H}{}_2\mathrm{O})=4.99-3.19=1.80\text{ g}$

$\begin{array}{rl}n(\mathrm{CuSO}{}_4)& =\frac{3.19}{159.62}=0.019985\text{ mol}\\ n(\mathrm{H}{}_2\mathrm{O})& =\frac{1.80}{18.02}=0.099889\text{ mol}\end{array}$

$x=\frac{n(\mathrm{H}{}_2\mathrm{O})}{n(\mathrm{CuSO}{}_4)}=\frac{0.099889}{0.019985}=4.998$

$x=5$, so the formula is $\mathrm{CuSO}{}_4\cdot 5\mathrm{H}{}_2\mathrm{O}$.

This is the calculation behind Finding an Empirical Formula, and the reason $x$ must be a whole number is that the water sits at definite positions in the crystal. A result of 4.6 does not mean a compound with 4.6 waters exists; it means some of the water never left.

5. Ammonium nitrate, $\mathrm{NH}{}_4\mathrm{NO}{}_3$, is used as a fertiliser and its value depends on how much nitrogen it delivers. Calculate its percentage composition.

Answer 5

Build the molar mass, keeping track of both nitrogens — one in the ammonium, one in the nitrate.

$M=2(14.01)+4(1.01)+3(16.00)=28.02+4.04+48.00=80.06\text{ g/mol}$

$\begin{array}{rl}\%\mathrm{N}& =\frac{28.02}{80.06}\times 100\%=35.00\%\\ \%\mathrm{H}& =\frac{4.04}{80.06}\times 100\%=5.046\%\\ \%\mathrm{O}& =\frac{48.00}{80.06}\times 100\%=59.96\%\end{array}$

Add them up as a check: $35.00+5.046+59.96=100.01\%$. The extra hundredth is rounding, not an error — the unrounded values sum to exactly 100%. If your total had come to 96% or 104%, you would have dropped or double-counted an atom, and the check would have caught it for free.

35.00% of the mass is nitrogen, which is the number a farmer actually buys. Two fertilisers at the same price per kilogram are not the same purchase if their nitrogen percentages differ, and that comparison is exactly what percentage composition is for.

6. A 1.000 g sample of a compound of iron and chlorine contains 0.3444 g of iron. Find its empirical formula.

Answer 6

$m(\mathrm{Cl})=1.000-0.3444=0.6556\text{ g}$

$\begin{array}{rl}n(\mathrm{Fe})& =\frac{0.3444}{55.85}=6.1665\times 10^{-3}\text{ mol}\\ n(\mathrm{Cl})& =\frac{0.6556}{35.45}=1.8494\times 10^{-2}\text{ mol}\end{array}$

Divide both by the smaller:

$\mathrm{Fe}:1.000\mathrm{Cl}:\frac{1.8494\times 10^{-2}}{6.1665\times 10^{-3}}=2.999$

The empirical formula is $\mathrm{FeCl}{}_3$, iron(III) chloride.

Sanity check with the chemistry you already have: iron(III) is $\mathrm{Fe}{}^{3+}$ and chloride is $\mathrm{Cl}{}^{-}$, so a one-to-three ratio is exactly what charge balance predicts. When a formula from data agrees with a formula from charges, both are probably right, and when they disagree it is worth finding out which one to trust before writing anything down.

7. Ethyne and benzene both have the empirical formula $\mathrm{CH}$. (a) Explain why the empirical formula alone cannot distinguish them. (b) Their molar masses are 26.04 g/mol and 78.12 g/mol. Find both molecular formulas. (c) What single measurement would you need to tell them apart?

Answer 7

(a) An empirical formula is a ratio, and a ratio is deliberately blind to size. One carbon per hydrogen describes a molecule with two atoms of each, one with six of each, and one with a hundred of each, equally well. Everything the empirical formula knows is preserved when you scale the molecule up, so nothing it knows can tell you the scale.

(b) $M(\mathrm{CH})=12.01+1.01=13.02\text{ g/mol}$

$\begin{array}{rl}\frac{26.04}{13.02}& =2.000\to \mathrm{C}{}_2\mathrm{H}{}_2\\ \frac{78.12}{13.02}& =6.000\to \mathrm{C}{}_6\mathrm{H}{}_6\end{array}$

(c) The molar mass, and nothing else is needed. That is the whole content of the relationship between empirical and molecular formulas: the ratio comes from composition, the multiplier comes from the molar mass, and neither measurement can be substituted for the other.

Worth being honest about the limits, though. Molar mass gives you the molecular formula and stops there. Two substances can share a molecular formula and be different compounds because the atoms are connected differently, and no mass measurement of any kind will separate those.

8. Three pieces of student work. Say what is wrong with each and give the correct result. (a) “My percentages were 48.63% C, 8.18% H, 43.19% O. I divided by the smallest and got C 1.50, H 3.00, O 1.00. I rounded the 1.50 up, so the empirical formula is $\mathrm{C}{}_2\mathrm{H}{}_3\mathrm{O}$.” (b) “My ratio came out Fe 1.00 : O 1.33, so I rounded to $\mathrm{FeO}$.” (c) “I found the empirical formula was $\mathrm{CH}{}_2$, so the molecular formula is $\mathrm{CH}{}_2$.”

Answer 8

(a) 1.50 is not a rounding error — it is a signal. A value that sits almost exactly halfway between two whole numbers is telling you that your unit is twice as big as it should be. Multiply every ratio by 2:

$\mathrm{C}:1.50\times 2=3\mathrm{H}:3.00\times 2=6\mathrm{O}:1.00\times 2=2$

The empirical formula is $\mathrm{C}{}_3\mathrm{H}{}_6\mathrm{O}{}_2$.

Two things went wrong at once. Rounding 1.50 to 2 changes the ratio by a third, which is far outside any experimental uncertainty. And the student multiplied only one of the three numbers, which changes the ratio rather than rescaling it — whatever you do, you do to all of them.

(b) Same signal, different fraction. 1.33 is close to $\frac{4}{3}$, so multiply both by 3:

$\mathrm{Fe}:3\mathrm{O}:4$

The empirical formula is $\mathrm{Fe}{}_3\mathrm{O}{}_4$, which is a real and common iron oxide — so rounding 1.33 down to 1 did not just lose precision, it named a different substance that was also sitting right there on the shelf. The fractions worth recognising on sight are 0.50 (multiply by 2), 0.33 and 0.67 (by 3), and 0.25 and 0.75 (by 4). Anything else — 1.15, say — is not a fraction to be cleared and is usually a sign that a mass or a molar mass went in wrong.

(c) The conclusion has no support. $\mathrm{CH}{}_2$ is the empirical formula, and the molecular formula is $(\mathrm{CH}{}_2){}_n$ for some whole number $n$ that the composition data cannot determine. It could be $\mathrm{C}{}_2\mathrm{H}{}_4$, $\mathrm{C}{}_3\mathrm{H}{}_6$, $\mathrm{C}{}_4\mathrm{H}{}_8$, and so on. To settle it you need the molar mass, exactly as in question 7. Assuming $n=1$ because no other information was given is not a conservative choice — it is an unsupported claim wearing the clothes of one.

Reference: Empirical and Molecular Formulas and Molar Mass and Composition. Doing it with a balance rather than with given data: Finding an Empirical Formula.

Curriculum connection

D2.4

determine the empirical formulae and molecular formulae of various chemical compounds, given molar masses and percentage composition or mass data [AI]

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D3.3

explain the relationship between the empirical formula and the molecular formula of a chemical compound

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