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 =
`(molar mass of element)/(Molar mass of the compound)* 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* 12.011)+(66* 1.008)+(4* 15.999) = 626.466 g/mol`

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

• For
  `Cu`

Mass percent of
`Cu` =
`(63.546)/(626.466)* 100`%

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

• For
  `C`

Mass percent of
`C` =
`(432.396)/(626.466)* 100`%

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

• For
  `H`

Mass percent of
`H` =
`(66.528)/(626.466)* 100`%

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

• For
  `O`

Mass percent of
`O` =
`(63.996)/(626.466)* 100`%

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