20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What be the percentage purity of magnsesium carbonate in the sample?

Options

A. 60
B. 84
C. 75
D. 96

Answer

2

Explanation

Percentage purity of magnesium carbonate in the sample can be calculate as following:
`"Percentage purity"=("mass of "MgCO_(3))/("mass of sample")xx100%`
The balanced chemical equation is
`MgCO_(3)(s) overset(Delta)(rarr) MgO(s)+CO_(2)(g)`
`24+12+16xx3`
`=84 u (24+16=40u)`
According to the above equation
`40 g MgO` needs `84 g MgCO_(3)`
`1.0 MgO` needs `84/40g MgCO_(3)`
`8.0 g MgO` needs `84/40xx8.0=16.8 g MgCO_(3)`
Thus
Percentage `purity=("mass of "MgCO_(3))/(mass_("sample"))xx100%`
`=(16.8 g)/(20.0 g)xx100%`
`=84%`
In the same question, they could have asked the percent yield of magnesium oxide:
`"Precent yield"=("actual yield")/("theoretical Yield")xx100%`
where the theoretical yield of a reaction is the amount of product predicted by the balanced equation when all of the reactant has reacted
`n_(MgCO_(3))=(mass_(MgCO_(3)))/("molar mass"_(MgCO_(3)))`
`20 g//84 g mol^(-1)`
`=0.238 mol`
According to the balanced equation, `1 mol e MgCO_(3)` yields `1 mol e MgO`. Thus, `0.238 mol Mg MgCO_(3)` will yield `0.238 mol MgO`.
`Mass_(MgO)=(n_(MgO))(Molar mass)`
`=(0.238 mol)(40 g mol^(-1))`
`=9.52 g`
Thus
percentage yield`=(8.00 g)/(9.52 g)xx100%`
`=84%`
Note that percentage purity of the reactant sample percentage yield of product are same