Question

Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce the metal oxides (MgO and CaO) and carbon dioxide (CO2). If 4.84 g of residue consisting of MgO and CaO remains when 9.66 g of dolomite is heated until decomposition is complete, what percentage by mass of the original sample was MgCO3?

Answer 1

Answer:

72.03 %

Explanation:

Total mass of dolomite = 9.66 g

Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol$

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

$Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol$

According to the reaction shown below:-

$MgCO_3\rightarrow MgO+CO_2$

$CaCO_3\rightarrow CaO+CO_2$

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = $\frac{x}{84.3139}\ mol$

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = $\frac{x}{84.3139}\times 40.3044 \ g$

Moles of CaO = $\frac{9.66 - x}{100.0869}\ mol$

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = $\frac{9.66 - x}{100.0869}\times 56.0774 \ g$

Given that total mass of the oxide = 4.84 g

Thus,

$\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84$

$\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84$

$-694.1618435x+45673.48749\dots =40843.38968\dots$

$x=\frac{4830.09780\dots }{694.1618435}$

$x=6.9582$

Thus, the mass of Magnesium carbonate = 6.9582 g

$\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100$

$\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%$