An unknown sample with a molecular mass of 180.0g is analyzed to yield 40% C, 6.7% H, and 53.3% O. What is the empirical formula and the molecular formula of this compound?

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An unknown sample with a molecular mass of 180.0g is analyzed to yield 40% C, 6.7% H, and 53.3% O. What is the empirical formula and the molecular formula of this compound?

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Answer 1 · 2016-04-19T15:46:06

$Empirical formula = C H_{2} O$ $"Empirical formula "=CH_2O$

$Molecular formula$ $"Molecular formula"$ $=$ $=$ $C_{6} H_{12} O_{6}$ $C_6H_12O_6$

Explanation:

We assume $100 \cdot g$ $100*g$ of compound and work out the atomic proportions:

$C : \frac{40.0 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}}$ $C: (40.0*g)/(12.011*g*mol^-1)$ $=$ $=$ $3.33$ $3.33$ $m o l$ $mol$ .

$H : \frac{6.7 \cdot g}{1.0794 \cdot g \cdot m o l^{- 1}}$ $H: (6.7*g)/(1.0794*g*mol^-1)$ $=$ $=$ $6.65$ $6.65$ $m o l$ $mol$ .

$O : \frac{53.3 .0 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}}$ $O: (53.3.0*g)/(15.999*g*mol^-1)$ $=$ $=$ $3.31$ $3.31$ $m o l$ $mol$ .

We divide thru by the lowest molar quantity to give us the empirical formula, $C H_{2} O$ $CH_2O$ , which is the simplest whole number ration defining constituent atoms in a species.

Now it is a fact that the molecular formula is always a multiple of the empirical formula.

Thus $Molecular formula = (Empirical formula) \times n$ $"Molecular formula " =" (Empirical formula")xxn$

$(12.011 + 2 \times 1.00794 + 15.999) \cdot g \cdot m o l^{- 1} \times n = 180.0 \cdot g \cdot m o l^{- 1}$ $(12.011+2xx1.00794+15.999)*g*mol^-1xxn = 180.0*g*mol^-1$

Clearly $n$ $n$ $=$ $=$ $6$ $6$ , and $molecular formula$ $"molecular formula"$ $=$ $=$ $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ .

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An unknown sample with a molecular mass of 180.0g is analyzed to yield 40% C, 6.7% H, and 53.3% O. What is the empirical formula and the molecular formula of this compound?

1 Answer

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