Question

In an experiment, 4.624 grams of a compound containing carbon, hydrogen and oxygen yieled 6.557 g of CO2 and 4.026 g of H2O in a combustion analysis. Determine the empirical formula of the compound.

What are the respective molar concentrations (molarity, M) of Fe3+ and I- ions obtained by dissolving 0.200 mol FeI3 in water and diluting 725 mL?

What volume of 3.25 M (NH4)2SO4 is needed in order to give 8.60 g of (NH4)2SO4 (132.14 g/mol).

Answer 1

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(6.557 g CO2) x (1 mol CO2/44.0 g CO2) x (1 mol C/1 mol CO2) x (12.011 g C/1
mol C) = 1.7899 g C
(4.026 g H2O) x (1 mol H2O/18.0 g H2O) x (2 mol H/1 mol H2O) x (1.008 g H/1
mol H) = 0.4509 g H
4.624 g – 1.7899 g C – 0.4509 g H = 2.3832 g O
(1.7899 g C) x (1 mol C/12.011 g C) = 0.1490 mol C
(0.4509 g H) x (1 mol H/1.008 g H) = 0.4473 mol H
(2.3832 g O) x (1 mol O/15.999 g O) = 0.1489 mol O
Divide through by 0.1489 to get CH₃O

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Express the concentration in mol per litre is the definition of molarity.
0.200mol in 725ml = 1000/725*0.2 = 0.275M
FeCl3 dissociate as:
FeCl3 ↔ Fe3+ + 3Cl-
1mol of FeCl3 gives 1 mol of Fe3+ ions
solution is 0.275M in FeCl3,

it must also be 0.275M in respect of Fe3+ ions,

and 3*0.275 = 0.825M in respect of Cl- ions.

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moles (NH4)2SO4 = 8.60 g/132.14 g/mol=0.0651
V = 0.0651/ 3.25 = 0.0200 L => 20.0 mL