Chemistry · General chemistry I · Worked example

Empirical formula from a fractional ratio

An iron oxide is $69.94 %$ iron by mass, and the rest is oxygen. What is its empirical formula?

69.94 % Fe, the rest O

Take a $100.00 g$ sample

Percent means per hundred, so $100.00 g$ of the oxide holds $69.94 g$ of iron. Oxygen is the only other element, so it makes up the rest.

100.00 g - 69.94 g = 30.06 g O

Convert iron to moles

Divide by the atomic mass of iron, $55.85 g / mol$ , written as $1 mol$ over $55.85 g$ so grams cancel. Keep four figures, because the next step looks for a fraction.

69.94 g Fe \times \frac{1 mol Fe}{55.85 g Fe} = 1.252 mol Fe

\begin{array}{c} 69.94 g Fe \times \frac{1 mol Fe}{55.85 g Fe} \\ = 1.252 mol Fe \end{array}

Convert oxygen to moles

Do the same with oxygen at $16.00 g / mol$ . There are more moles of oxygen than of iron, even though there are fewer grams, because an oxygen atom is the lighter of the two.

30.06 g O \times \frac{1 mol O}{16.00 g O} = 1.879 mol O

\begin{array}{c} 30.06 g O \times \frac{1 mol O}{16.00 g O} \\ = 1.879 mol O \end{array}

Divide by the smaller amount

Iron has the smaller amount, so it becomes $1$ . Oxygen comes out at $1.501$ . That is not close enough to $1$ or to $2$ to round, and it sits right on one and a half.

\frac{1.879 mol O}{1.252 mol Fe} = 1.501

Clear the half

Atoms come whole, so multiply both numbers in the ratio by $2$ . Iron becomes $2$ and oxygen $3.002$ , which is $3$ . The empirical formula is ${Fe}_{2} O_{3}$ .

Mole ratios in $100.00 g$ of the oxide
Element	Moles	$\div 1.252$	$\times 2$
Fe	$1.252$	$1.000$	$2$
O	$1.879$	$1.501$	$3$

Check against the data

Work backward from the formula. ${Fe}_{2} O_{3}$ has a molar mass of $2 (55.85) + 3 (16.00)$ $= 159.70 g / mol$ , of which $111.70 g$ is iron. That is $69.94 %$ , the percentage the problem gave.

\frac{111.70 g Fe}{159.70 g} \times 100 % = 69.94 %

Result

The empirical formula is ${Fe}_{2} O_{3}$ , iron(III) oxide.

{Fe}_{2} O_{3}

Your turn

1. One like the example

Another iron oxide is $72.36 %$ iron and $27.64 %$ oxygen. What is its empirical formula?

Show the answer and explanation

${Fe}_{3} O_{4}$ .

In $100 g$ there are $72.36 \div 55.85$ $= 1.296 mol$ of iron and $27.64 \div 16.00$ $= 1.728 mol$ of oxygen. $1.728 \div 1.296 = 1.333$ , a third over $1$ . Multiplying $1$ : $1.333$ by $3$ gives $3$ : $4$ , so the formula is ${Fe}_{3} O_{4}$ , the magnetic oxide.

\begin{array}{l} m = 100.00 g \\ 72.36 g Fe \times \frac{1 mol Fe}{55.85 g Fe} = 1.296 mol Fe \\ 27.64 g O \times \frac{1 mol O}{16.00 g O} = 1.728 mol O \\ \frac{1.728 mol O}{1.296 mol Fe} = 1.333 \end{array}

\begin{array}{c} m = 100.00 g \\ 72.36 g Fe \times \frac{1 mol Fe}{55.85 g Fe} \\ = 1.296 mol Fe \\ 27.64 g O \times \frac{1 mol O}{16.00 g O} \\ = 1.728 mol O \\ \frac{1.728 mol O}{1.296 mol Fe} = 1.333 \end{array}

2. A variation

Vitamin C is $40.92 %$ carbon, $4.58 %$ hydrogen and $54.50 %$ oxygen by mass. What is its empirical formula?

Show the answer and explanation

$C_{3} H_{4} O_{3}$ .

In $100 g$ there are $3.407 mol$ of carbon, $4.54 mol$ of hydrogen and $3.406 mol$ of oxygen. Carbon and oxygen are equal, and $4.54 \div 3.406 = 1.33$ for hydrogen. The ratio $1$ : $1.33$ : $1$ has a third in it, so multiply by $3$ to get $3$ : $4$ : $3$ , $C_{3} H_{4} O_{3}$ .

\begin{array}{l} m = 100.00 g \\ 40.92 g C \times \frac{1 mol C}{12.01 g C} = 3.407 mol C \\ 4.58 g H \times \frac{1 mol H}{1.008 g H} = 4.54 mol H \\ 54.50 g O \times \frac{1 mol O}{16.00 g O} = 3.406 mol O \\ \frac{4.54 mol H}{3.406 mol O} = 1.33 \end{array}

\begin{array}{c} m = 100.00 g \\ 40.92 g C \times \frac{1 mol C}{12.01 g C} \\ = 3.407 mol C \\ 4.58 g H \times \frac{1 mol H}{1.008 g H} \\ = 4.54 mol H \\ 54.50 g O \times \frac{1 mol O}{16.00 g O} \\ = 3.406 mol O \\ \frac{4.54 mol H}{3.406 mol O} = 1.33 \end{array}

3. Somewhere new

A hydrocarbon burns completely to $0.4401 g$ of ${CO}_{2}$ and $0.2403 g$ of $H_{2} O$ . What is its empirical formula?

Show the answer and explanation

$C_{3} H_{8}$ .

Each ${CO}_{2}$ carries one carbon atom from the fuel, so $0.4401 g$ of ${CO}_{2}$ means $0.01000 mol C$ . Each $H_{2} O$ carries two hydrogen atoms, so $0.2403 g$ of water means $0.02667 mol$ of H. $0.02667 \div 0.01000$ $= 2.667$ , two and two thirds. Tripling $1$ : $2.667$ gives $3$ : $8$ , which is $C_{3} H_{8}$ .

\begin{array}{l} m = 0.4401 g \\ 0.4401 g {CO}_{2} \times \frac{1 mol {CO}_{2}}{44.01 g {CO}_{2}} \times \frac{1 mol C}{1 mol {CO}_{2}} = 0.01000 mol C \\ 0.2403 g H_{2} O \times \frac{1 mol H_{2} O}{18.02 g H_{2} O} \times \frac{2 mol H}{1 mol H_{2} O} = 0.02667 mol H \\ \frac{0.02667 mol H}{0.01000 mol C} = 2.667 \end{array}

\begin{array}{c} m = 0.4401 g \\ 0.4401 g {CO}_{2} \\ \times \frac{1 mol {CO}_{2}}{44.01 g {CO}_{2}} \times \frac{1 mol C}{1 mol {CO}_{2}} \\ = 0.01000 mol C \\ 0.2403 g H_{2} O \\ \times \frac{1 mol H_{2} O}{18.02 g H_{2} O} \times \frac{2 mol H}{1 mol H_{2} O} \\ = 0.02667 mol H \\ \frac{0.02667 mol H}{0.01000 mol C} = 2.667 \end{array}

Keep exploring

Open the data in Composition & formulas and change the iron to $72.36 %$ : the ratio becomes $1$ : $1.333$ , and tripling it gives ${Fe}_{3} O_{4}$ .

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Take a 100.00 g100.00 g sample

Convert iron to moles

Convert oxygen to moles

Divide by the smaller amount

Clear the half

Check against the data

Result

Your turn

1. One like the example

Fe3O4Fe3​O4​.

2. A variation

C3H4O3C3​H4​O3​.

3. Somewhere new

C3H8C3​H8​.

Keep exploring

Study it, then try it

Take a $100.00 g$ sample

${Fe}_{3} O_{4}$ .

$C_{3} H_{4} O_{3}$ .

$C_{3} H_{8}$ .