A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liberated. What percentage of BaCO3 did the sample contain? Assume all the carbon originally present is converted to carbon dioxide.

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O = 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2) / (molar mass) = 0.234/44.011 = 0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass) / (molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34 = 0.00532 (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3 = (mass of BaCO3) / (mass of sample) * 100

= 0.242 / (0.846) * 100

= 28.605%