Question

Determine the mass percent of each of the elements in the fungicide copper(ii) oleate, cu(c18h33o2)2.

Answer 1

The mass percent is determined by the formula:

mass percent = $\frac{molar mass of element}{Molar mass of the compound}\times 100$    -(1)

Given compound is copper(II) oleate, $Cu(C_{18}H_{33}O_2)_2$

The atomic mass of the elements present in copper(II) oleate, $Cu(C_{18}H_{33}O_2)_2$ are:

Atomic mass of $Cu$ = $63.546 u$

Atomic mass of $C$ = $12.011 u$

Atomic mass of $H$ = $1.088 u$

Atomic mass of $O$ = $15.999 u$

The molar mass of $Cu(C_{18}H_{33}O_2)_2$ = $63.546+(36\times 12.011)+(66\times 1.008)+(4\times 15.999) = 626.466 g/mol$

Using formula (1), the mass percent of each element is calculated as:

• For $Cu$

Mass percent of $Cu$ = $\frac{63.546}{626.466}\times 100$%

Mass percent of $Cu$ = $10.14$ %.

• For $C$

Mass percent of $C$ = $\frac{432.396}{626.466}\times 100$%

Mass percent of $C$ = $69.02$ %.

• For $H$

Mass percent of $H$ = $\frac{66.528}{626.466}\times 100$%

Mass percent of $H$ = $10.62$ %.

• For $O$

Mass percent of $O$ = $\frac{63.996}{626.466}\times 100$%

Mass percent of $O$ = $10.21$ %.