Question

The percent composition by mass of an unknown compound with a molecular mass of 90.078 u is 40.002% C, 6.7135% H, and 53.284% O. Determine the compound's empirical and molecular formulas.

Answer 1

Answer:

7u

Explanation:

Answer 2

The empirical formula of the unknown compound is $C_1H_2O_1$.

The molecular formula of the unknown compound is $C_3H_6O_3$.

Given:

An unknown compound with a molecular mass of 90.078 u.

The unknown compound has 40.002% C, 6.7135% H, and 53.284% O

To find:

The empirical and molecular formula of the unknown compound.

Solution:

Consider in 100 grams of an unknown compound

Amount of carbon in 100 grams of compound = 40.002% of 100 g

$=\frac{40.002 }{100}\times 100 g=40.002 g$

Moles of carbon :

= $\frac{40.002 g}{12.0107 g/mol}=3.3305 mol$

Amount of hydrogen in 100 grams of compound = 6.7135% of 100 g

$=\frac{ 6.7135}{100}\times 100 g= 6.7135g$

Moles of hydrogen:

=$\frac{6.7135g}{1.00784g/mol}=6.6613mol$

Amount of oxygen in 100 grams of compound = 53.284% of 100 g

$=\frac{ 53.284}{100}\times 100 g= 53.284g$

Moles of oxygen:

=$\frac{ 53.284g}{15.999g/mol}=3.3305 mol$

Divide the moles of every element with the least number of moles of an element to obtain a number of atoms of each element in the empirical formula of the unknown compound

Let the empirical formula of compound be : $C_xH_yO_z$

• For carbon atoms:

$x=\frac{3.3305 mol}{3.3305 mol}=1$

• For hydrogen atoms:

$y=\frac{6.6613mol}{3.3305 mol}=2$

• For oxygen atoms:

$z=\frac{3.3305 mol}{3.3305 mol}=1$

The empirical formula of the unknown compound =$C_1H_2O_1$

The empirical mass of the unknown compound:

$E.M=1\times 12.0107 u+ 2\times 1.00784u+1\times 15.999 u\\E.M=30.02538 u$

The molecular formula of the unknown compound =$C_{1n}H_{2n}O_{1n}$

The molecular mass of an unknown compound = M.M= 90.078 u

$n=\frac{M.M}{E.M}\\=\frac{90.078 u}{30.02538 u}=3$

The molecular formula of the compound:

$=C_{1\times 3}H_{2\times 3}O_{1\times 3}=C_3H_6O_3$

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