Find the percentage of SrO in a sample of SrCO_3 and Li_2CO_3.

Question

A sample of mixed carbonates, SrCO_3 (147.63 g/mol) and Li_2CO_3 (73.89 g/mol), weighed 0.5310 grams. If this sample is titrated with acid, it is found that 20.10 mL of 0.5010 M HCl are required to reach the endpoint. What is the percentage of SrO (103.62 g/mol) in the sample? SrCO_3 and Li_2CO_3 are the only components in the sample.

Answer 1

first find the moles of the only thing we can, HCl:

0.02010 litres @ 0.5010 mol / litre = 0.01007 moles of HCl

2 HCl's react with either 1 SrCO3 or 1 Li2CO3, so....

0.01007 moles of HCl = 0.005035 moles of the carbonate salts

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since SrCO3 & Li2CO3 add up to 0.5310 grams, i'll pick one to be x:

SrCO3 = x grams , so

Li2CO3 = 0.5310 -x grams

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now I am going to convert those grams into moles, knowing that they total 0.005035 moles, & solve for the grams of SrCO3 @ the same time:

SrCO3 moles & Li2CO3 moles = 0.005035 moles

(X grams) / (147.63 g/mol) & (0.5310g-X) / (73.89g/mol) = 0.005035 moles

0.006774X & 0.007186 - 0.01353X = 0.005035

- 0.006760X = - 0.002151

X = 0.3182 grams of SrCO3

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now lets convert the grams of SrCO3 into grams of SrO:

0.3182 g SrCO3 @ 103.62 g SrO / 147.63 g SrCO3 = 0.2233 grams s****.>
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lastly , the %

0.2233 grams SrO / 0.5310 g sample (100) = 42.06 %

your answer is 42.06% SrO