A compound having an empirical formula of $C_{3} H_{4} O_{3}$ $C_3H_4O_3$ has a mass of 180 g/mol. What is the molecular formula?

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A compound having an empirical formula of $C_{3} H_{4} O_{3}$ $C_3H_4O_3$ has a mass of 180 g/mol. What is the molecular formula?

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Answer 1 · 2016-06-04T08:46:37

$Molecular formula$ $"Molecular formula"$ $=$ $=$ $C_{6} H_{8} O_{6}$ $C_6H_8O_6$

Explanation:

The molecular formula is always a whole number multiple of the empirical formula:

i.e. $Molecular formula$ $"Molecular formula"$ $=$ $=$ $(Empirical formula) \times n$ $"(Empirical formula)"xxn$

So we need to solve for $n$ $n$ .

Thus $180 \cdot g \cdot m o l^{- 1}$ $180*g*mol^-1$ $=$ $=$ $n \times (3 \times 12.011 + 4 \times 1.0079 + 3 \times 15.999) \cdot g \cdot m o l^{- 1}$ $nxx(3xx12.011+4xx1.0079+3xx15.999)*g*mol^-1$
$=$ $=$ $n \times (88.0) \cdot g \cdot m o l^{- 1}$ $nxx(88.0)*g*mol^-1$

$n ≅ 2$ $n~=2$ . Thus $Molecular formula$ $"Molecular formula"$ $=$ $=$ $C_{6} H_{8} O_{6}$ $C_6H_8O_6$ .

Of course, the ratio is out a little bit. Such errors are common in the determination of such formulae.

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A compound having an empirical formula of $C_{3} H_{4} O_{3}$ $C_3H_4O_3$ has a mass of 180 g/mol. What is the molecular formula?

1 Answer

Explanation:

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A compound having an empirical formula of C_3H_4O_3C3H4O3C_3H_4O_3 has a mass of 180 g/mol. What is the molecular formula?

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Explanation:

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A compound having an empirical formula of $C_{3} H_{4} O_{3}$ $C_3H_4O_3$ has a mass of 180 g/mol. What is the molecular formula?