need help with quantitative analysis problems Exam A o.2121g Sample of an argani

Question

need help with quantitative analysis problems

Exam A o.2121g Sample of an arganic campound was bunted in a Stream of oxygen,and the Coz preduced was celleceel in a sclution of barium kydroxide aleulate theof arhon CAtwt 12.. aig) in the sample ip o.6006 g of Baco Cia734ylmal) was Pormed 2) | A o. 8720 g Sample af a mixture consisting solely?Sodium bro male (102. 89Hglna?and Potassium bronale C i?.ooegld) yields I 5osaof siler bromide C187.772glmety Glonlat the % ee each. Salt inthe Sample 3)A 50.00 mL Solutional 0.05oomformic acid.(ka l'80x12)-- was titrated with o. looo M KOH Celeulate the pHof tRe Soution aPter the adit495 and 25.05 mLof KoH 4) A 25.00 ml Solution that is o. Is Min HCl and o12M ifamic -ty e Sa rthe Calcalate the pte the Solution aleer -the addition f 5-0o, 30.00 and HooamLef HCL

Explanation / Answer

1)

BaOH2 + CO2 = BaCO3 + H2O

1 mole of BaOH2 reacts with 1 mole of CO2 gives 1 mole of BaCO3

Here we have 0.6006 g of BaCO3

We need to find the number of moles of BaCO3 in 0.6006 g of BaCO3

Given the Molecular mass of BaCO3 is 197.34 g / mol

i.e)

197.34 g = 1 mol

0.6006 g of BaCO3 = 0.6006 / 197.34 mol

= 0.00304 mol

So 0.6006 g of BaCO3 has 0.00304 moles

Which implies 0.00304 moles of CO2 reacted with 0.00304 moles of BaOH2

So the amount of carbon present in the organic compound is 0.00304 moles.

Now,

the mass of carbon is

= 0.00304 mol x 12.011 g / mol [Given molar mass of CO2]

= 0.03651 g of carbon

The total mass of the organic compound is 0.2121 g

The percentage of carbon is [0.03651 / 0.2121 ] x 100 = 17.21 %

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