Question

1. a compound consists of 52.14 % carbon, 13.13% of hydrogen, and 34.73% of oxygen by mass. (a) what is the empirical formula? (b) what is the molecular formula if the molar mass of the compound is 138.204 g/mol? element mass (g) atomic mass mole = mass / atomic mass mole ratio = mole / smallest mole c 52.14 12.01 4.34 2 h 13.13 1.008 13.02 6 o 34.73 16.00 2.17 1 empirical formula: c₂h₆o

Explanation:

Step1: Determine the empirical - formula

Assume we have 100 g of the compound. So, we have 52.14 g of C, 13.13 g of H, and 34.73 g of O.
Moles of C: $n_{C}=\frac{52.14\ g}{12.01\ g/mol}\approx4.34\ mol$
Moles of H: $n_{H}=\frac{13.13\ g}{1.01\ g/mol}\approx13.0\ mol$
Moles of O: $n_{O}=\frac{34.73\ g}{16.00\ g/mol}\approx2.17\ mol$
The mole - ratio of C : H : O is $\frac{4.34}{2.17}:\frac{13.0}{2.17}:\frac{2.17}{2.17}\approx2:6:1$. So the empirical formula is $C_{2}H_{6}O$.

Step2: Calculate the empirical - formula mass

The empirical - formula mass of $C_{2}H_{6}O$ is $2\times12.01 + 6\times1.01+1\times16.00=46.08\ g/mol$.

Step3: Determine the multiple 'n'

The molar mass of the compound is given as 138.204 g/mol.
$n=\frac{\text{molar mass}}{\text{empirical - formula mass}}=\frac{138.204\ g/mol}{46.08\ g/mol}\approx3$.

Step4: Find the molecular formula

Multiply the sub - scripts in the empirical formula by 'n'.
The molecular formula is $(C_{2}H_{6}O)_{3}=C_{6}H_{18}O_{3}$.

Answer:

$C_{6}H_{18}O_{3}$