Question

A carbon containing compound was treated chemically to convert all its carbon into SrCO3. A 31.23 g sample of the compound gave 1.203 x 102 g SrCO3. a) What is the percentage of carbon in the compound? b) If there are three carbon atoms in a molecule of the compound, what is the molar mass

Answer 1

1 mole of SrCO3 contains 1 mole of C

147.6g of SrCO3 contains 12g of C

1.203*10^2g of SrCO3 contains = 12*1.203*10^2/147.6   =9.78g of C

percentage of carbon (C%) in the sample = 9.78*100/31.23   = 31.32%

b.

1 mole of SrCO3 contains 1 mole of C

147.6g of SrCO3 contains 1mole of C

1.203*10^2g of SrCO3 contains   = 1mole*1.203*10^2g/147.6g = 0.815moles of C

three carbon atoms in a molecules of compound = 0.815/3    = 0.272moles of C

molar mass of sample = mass of sample/no of moles

                                     = 31.23/0.272 = 114.8g/mole >>>>>answer