A molecule with molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. What are the empirical and molecular formulas of the molecule?

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A molecule with molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. What are the empirical and molecular formulas of the molecule?

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Answer 1 · 2014-06-20T02:23:32

The empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆.

Assume we have 100 g of the compound.

Then we have 40.00 g of C, 6.72 g of H, and 53.28 g of O.

Moles of C = 40.00 g C × $\frac{1 mol C}{12.01 g C}$ $(1"mol C")/(12.01"g C")$ = 3.331 mol C

Moles of H = 6.72 g H × $\frac{1 mol H}{1.008 g H}$ $(1" mol H")/(1.008"g H")$ = 6.67 mol H

Moles of O = 53.28 g O × $\frac{1 mol O}{16.00 g O}$ $(1" mol O")/(16.00"g O")$ = 3.330 mol H

Moles of C:Moles of H:Moles of O = 3.331:6.67:3.330=1.000:2.00:1 ≈ 1:2:1

The empirical formula is CH₂O.

The empirical formula is the simplest formula of a compound. The actual formula is an integral multiple of the empirical formula.

If the empirical formula is CH₂O, the actual formula is ${(CH₂O)}_{n}$ $"(CH₂O)"_n$ or $C_{n} H_{2 n} O_{n}$ $C_nH_(2n)O_n$ , where $n$ $n$ = 1, 2, 3, … Our job is to determine the value of $n$ $n$ .

The empirical formula mass of CH₂O is 30.03 u. The molecular mass of 180.18 u must be some multiple of this number.

$n = \frac{180.18 u}{30.03 u}$ $n = (180.18" u")/(30.03" u")$ = 6.000 ≈ 6

∴ The molecular formula = $C_{n} H_{2 n} O_{n}$ $C_nH_(2n)O_n$ = C₆H₁₂O₆.

Hope this helps.

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A molecule with molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. What are the empirical and molecular formulas of the molecule?

1 Answer

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