CALCULATING A PERCENT COMPOSITION FROM A FORMULA Glucose, or blood sugar, has the molecular formula [latex]C_{6}H_{12}O_{6}[/latex]. What is the empirical formula, and what is the percent composition of glucose?

The empirical formula is found by reducing the subscripts in the molecular formula to their smallest whole-number values. In this case, dividing the subscripts by 6 reduces [latex]C_{6}H_{12}O_{6} to CH_{2}O[/latex]. The percent composition of glucose can be calculated either from the molecular formula or from the empirical formula. Using the molecular for C:H:O mole ratio of 6:12:6 can be converted into a mass ratio by assuming that we have 1 mol of compound and carrying out mole-to-gram conversions: [latex]1 \cancel{mol glucose} × \frac{6 \cancel{mol C}}{1 \cancel{mol glucose}} × \frac{12.0 g C}{1 \cancel{mol C}}[/latex] = 72.0 g C [latex]1 \cancel{mol glucose} × \frac{12 \cancel{mol H}}{1 \cancel{mol glucose}} × \frac{1.01 g H}{1 \cancel{mol H}}[/latex] = 12.1 g H [latex]1 \cancel{mol glucose} × \frac{6 \cancel{mol O}}{1 \cancel{mol glucose}} × \frac{16.0 g O}{1 \cancel{mol O}}[/latex] = 96.0 g O Dividing the mass of each element by the total mass, and multiplying by 100%, gives the percent composition. Note that the sum of the mass percentages is 100%. Total mass of 1 mol glucose = 72.0 g + 12.1 g + 96.0 g = 180.1 g % C = [latex]\frac{72.0 g C}{180.1 g}[/latex] × 100% = 40.0% % H = [latex]\frac{12.1 g H}{180.1 g}[/latex] × 100% = 6.72% % O = [latex]\frac{96.0 g O}{180.1 g}[/latex] × 100% = 53.3%

Question 3.16: CALCULATING A PERCENT COMPOSITION FROM A FORMULA Glucose, or...

General Chemistry: Atoms First [1015060]
Chemistry

CALCULATING A PERCENT COMPOSITION FROM A FORMULA

Glucose, or blood sugar, has the molecular formula $C_{6}H_{12}O_{6}$ . What is the empirical formula, and what is the percent composition of glucose?

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The empirical formula is found by reducing the subscripts in the molecular formula to their smallest whole-number values. In this case, dividing the subscripts by 6 reduces $C_{6}H_{12}O_{6} to CH_{2}O$ .

The percent composition of glucose can be calculated either from the molecular formula or from the empirical formula. Using the molecular for C:H:O mole ratio of 6:12:6 can be converted into a mass ratio by assuming that we have 1 mol of compound and carrying out mole-to-gram conversions:

$1 \cancel{mol glucose} × \frac{6 \cancel{mol C}}{1 \cancel{mol glucose}} × \frac{12.0 g C}{1 \cancel{mol C}}$ = 72.0 g C

$1 \cancel{mol glucose} × \frac{12 \cancel{mol H}}{1 \cancel{mol glucose}} × \frac{1.01 g H}{1 \cancel{mol H}}$ = 12.1 g H

$1 \cancel{mol glucose} × \frac{6 \cancel{mol O}}{1 \cancel{mol glucose}} × \frac{16.0 g O}{1 \cancel{mol O}}$ = 96.0 g O

Dividing the mass of each element by the total mass, and multiplying by 100%, gives the percent composition. Note that the sum of the mass percentages is 100%.

Total mass of 1 mol glucose = 72.0 g + 12.1 g + 96.0 g = 180.1 g

% C = $\frac{72.0 g C}{180.1 g}$ × 100% = 40.0%

% H = $\frac{12.1 g H}{180.1 g}$ × 100% = 6.72%

% O = $\frac{96.0 g O}{180.1 g}$ × 100% = 53.3%