Ch107_Lec_Stoichiometry Exercises

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107
Fundamentals of Chemistry
Fall 2009

Lecture Notes: Stoichiometry Examples
© R. Paselk 2005

Worked Examples

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*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
Chem 107	Fundamentals of Chemistry	Fall 2009
Lecture Notes: Stoichiometry Examples	© R. Paselk 2005
	Worked Examples
	Use Back Button to Return to Notes

A compound of mercury, oxygen, and nitrogen gave an analysis of 61.80% mercury and 8.63% nitrogen by weight. What is the empirical formula for this compound?

First need to find the amount of oxygen: 100% - 61.80% - 8.63% = 29.57%

Next we need to find the number of moles. Easiest to assume 100 g total, and find moles of each:

Hg: (61.80g)/(200.6g/mole) = 3.081 x 10^-1moles

N: (8.63g)/(14.01g/mol) = 6.16 x 10^-1moles

O: (29.57g)/(16.00g/mol) = 1.848 moles

So formula = Hg_0.3081N_0.616O_1.848

But we want whole number ratios, so divide each coefficient by the smallest:

Hg_{0.3081/0.3081}N_0.616/0.3081O_1.848/0.3081 to get:

Hg₁N_1.999O_5.998, and rounding off

= HgN₂O₆

A molecule is found to have an empirical formula of C₃H₅ by elemental analysis and a MW of 85 by freezing point depression. What is its molecular formula?

First we need to determine the formula weight = 3 (12.01) + 5 (1.008) = 41.07,

Then we divide the MW by the FW: 85/41 = 2.07.

Now we know the MW is not terribly precise, so we use it as a decision number, that is the the MW = FW, or 2 x FW, or 3 x FW etc. Obviously ours is 2 x, so:

Molecular formula = C₆H₁₀

Yield example: How many g of water will be produced during the complete combustion of 475.5 g of natural gas (methane)?

First we need to find the number of moles of CH₄ = 475.5g/16.04 g/mole = 29.64 moles.

Next we need to write the chemical equation for this process:

CH₄ + 2 O₂

CO₂ + 2 H₂O So we get 2 moles of water for every mole of methane

= (29.64 moles) x 2 = 59.28 moles

(59.28 moles)(18.02 g/mole) = 1,068 g

Limiting example: Consider the reaction:

Fe₃O₄ + 4 C

3 Fe + 4 CO

What is the maximum mass of Fe which could be made from 115.0 g Fe₃O₄ of and 24.00 g C?

The trick here is to find the maximum amount of iron which could be made from each reactant. The lesser amount will then be the max possible:

C: (3 mol Fe/ 4 mol C)(24.00 g C/ 12.01 g C/mol C) = 1.499 mole

Fe₃O₄: (3 mol Fe/ mol Fe₃O₄)(115.0 gFe₃O₄/231.6 g Fe₃O₄/mol Fe₃O₄) = 1.490 mole

\ Fe₃O₄ limits, can only make 1.490- moles.

So mass is then= (1.490 moles) (55.85 g/mole) = 83.22 g Fe

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Last modified 28 October 2005

Humboldt State University ® Department of Chemistry

Richard A. Paselk