Question

A 1.850 g mixture of SrCO3 and SrO is heated. The SrCO3 decomposes to SrO and CO2. What was the mass percentage of SrCO3 in the mixture if the mass after heating is 1.445 g?

Answer 1

decomposition of SrCO3 to SrO and CO2 =change in mass

 moles of CO2 =(1.850 g - 1.445 g). 
Mass of CO2 in mixture: 1.850-1.445 = 0.405g
0.405g/44.01 g/mol CO2 = 0.0092 moles CO2.
ratio of CO2 to SrO in SrCO3 is 1:1
 mass ratio = 1.358/1.850 = 0.7341, 
or 73.41% SrCO3.
hope this helps

Answer 2

Answer:

The percent composition of strontium carbonate in the original sample was 73.43 %

Explanation:

Heating the sample will consume all strontium carbonate,  SrCO₃. In this way strontium oxide SrO is produced; and carbon dioxide, CO₂ is emitted by the following reaction:

SrCO ₃ →  SrO  + CO ₂

This is a decomposition reaction.

 Then, the difference between the initial mass of the sample and the final mass of the sample is equal to the mass of carbon dioxide produced by the reaction:

1.850 g -1.445 g= 0.405 g

You know that C has 12 g/mol and O 16 g/mol, so the molar mass of CO₂ is 12 g/mol +2*16 g/mol= 44 g/mol

Then, a rule of three is applied to know the amount of moles that represent 0.405 g of CO₂: if 44 g is 1 mole of the compound, 0.405 g how many moles do they represent?

$moles of CO_{2}=\frac{0.405 g* 1 mole}{44g}$

moles of CO₂=0.009204

By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), in the decomposition reaction written above you can see that each mole of strontium carbonate produces one mole of carbon dioxide. Then, applying rule of three it is possible to deduce that 0.009204 moles of CO₂ produce 0.009204 moles of SrCO₃.

Knowing that:

• Sr: 87.62 g/mol
• C: 12 g/mol
• O: 16 g/mol

Then the molar mass of SrCO₃ is 87.62 g/mol + 12 g/mol + 3*16 g/mol= 147.62 g/mol

Then, a rule of three is applied to know the amount of mass that represent 0.009204 moles of SrCO₃: if 147.62 g is 1 mole of the compound, 0.009204 moles how many mass do they represent?

$mass of SrCO_{3}=\frac{0.009204moles*147.62 g}{1 mole}$

mass of SrCO₃=1.3586 g

Then, the percent composition of strontium carbonate in the original sample was:

$percentof strontium carbonate=\frac{1.3586g}{1.850g} *100$

percentof strontium carbonate=73.43 %