Question

An organic compound weighs 0.316 g after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g. of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic mass: Ba = 137, S = 32, O = 16, C = 12, H = 1)

Answer 1

Hint: The mass percentage of the atom in a compound is calculated by dividing the mass of the atom by the total mass of the chemical compound multiplied with 100. The molecular mass, mass and number of moles is related by the formula where number of moles is calculated by dividing mass by the molecular mass.

Complete step by step answer:
Given,
The mass of organic compound is 0.316 g.
Mass of barium sulphate is 0.466 g.
Atomic mass of Ba is 137
Atomic mass of O is 16
Atomic mass of carbon is 12
Atomic mass of hydrogen is 1.
Atomic mass of sulphur is 32.
The mass of sulphur in 0.466 g of barium sulphate is calculated as shown below.
${m_S} = \dfrac{{{m_{BaS{O_4}}} \times {A_S}}}{{{M_{BaS{O_4}}}}}$
${m_S}$is the mass of sulphur
${A_S}$ is the atomic mass of sulphur.
${M_{BaS{O_4}}}$ is the molecular weight of barium sulphate.
The molecular weight of barium sulphate is 233 g/mol.
The mass of sulphur is calculated is shown below.
\[ \Rightarrow {m_S} = \dfrac{{0.466 \times 32}}{{233}}\]
\[ \Rightarrow {m_S} = 0.064g\]
The mass percentage is calculated by the formula shown below.
$mass\% = \dfrac{{mass\;of\;sulphur}}{{Total\;mass\;of\;compound}} \times 100$
To calculate the mass % of sulphur, substitute the value in the above equation.
$ \Rightarrow mass\% = \dfrac{{0.064g}}{{0.316}} \times 100$
$ \Rightarrow mass\% = 20.253$%
Therefore, the mass % of the sulphur in the compound is 20.253 %.

Note:
Don’t get confused by the given statement as the unknown chemical compound is the reactant which combines with nitric acid and barium nitrate to form the product barium sulphate and we need to find the percentage of sulphur in the unknown chemical compound whose weight is given.