Ch107_Lec_Concentration Exercises

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107
Fundamentals of Chemistry
Fall 2009

Lecture Notes: Concentration 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: Concentration Examples	© R. Paselk 2005
	Worked Examples
	Use Back Button to Return to Notes

% quantities:

What is the concentration of iron in iron(II) sulfate.

First need chemical formula = FeSO₄

So formula weight = 55.85 + 32.07 + 4(16.00) = 151.92

and % = (55.85/151.92) x 100% = 36.76%

How many mg of Fe are there in a 312 mg tablet of iron(II) sulfate?

For this problem we can use the fraction of iron found above, 36.76% = 0.3676, so

0.3676 x 312 mg = 114.7 mg = 115 mg

Masses and particles:

What is the weight of 0.25 mole of:
- KBr?

0.25 mole x (39.10 g/mole + 79.90 g/mole) = 29.75 g = 3.0 x 10 g

glucose?

First need formula = C₆H₁₂O₆

so weight = 0.25 mole x (6 [12.01]g/mole + 12 x [1.008 g/mole] + 6 [16.00] g/mole) = 45.04 g = 45 g

ethanol?

First need formula = C₂H₅OH

so weight = 0.25 mole x (2 [12.01]g/mole + 6 x [1.008 g/mole] + 16.00 g/mole) = 11.52 g = 12 g

How many moles are there in 23.5 g of:
- sucrose

First need formula = C₁₂H₂₂O₁₁

so MW = (12 [12.01]g/mole + 22 x [1.008 g/mole] + 11 [16.00 ] g/mole) = 342.30 g/mole

and moles = (23.5 g)/(342.30 g/mole) = 6.865 x 10^-2mole = 6.87 x 10^-2mole.

NaCl

FW = (22.99 g/mole + 35.45 g/mole) = 58.44 g/mole

and moles = (23.5 g)/(58.44g/mole) = 4.021 x 10^-1mole = 4.02 x 10^-1mole.

propane (C₃H₈)

MW = (3 [12.01]g/mole + 9 [1.008 g/mole]) = 45.102 g/mole

and moles = (23.5 g)/(45.102 g/mole) = 5.21 x 10^-1mole.

How many atoms are there in 1.00 g of
- KBr?

First need FW = (39.10 g/mole + 79.90 g/mole) = 119.0 g/mole

then find moles = (1.00 g)/(119.0 g/mole) = 8.403 x10^-3mole

then multiply by Avogadro's Number: (8.403 x10^-3mole) x (6.022 x 10^-3 atoms/mole) = 5.061 x10²¹ KBr "units"

But - there are two atoms/formula, So 1.01 x10²² atoms

glucose?

First need formula = C₆H₁₂O₆

so MW = (6 [12.01]g/mole + 12 x [1.008 g/mole] + 6 [16.00] g/mole) = 180.16 g/mole

then find moles = (1.00 g)/(180.16 g/mole) = 5.551 x10^-3mole

then multiply by Avogadro's Number: (5.551 x10^-3mole) x (6.022 x 10^-3 atoms/mole) = 3.343 x10²¹ glucose molecules

But - there are 24 atoms/formula, So 8.02 x10²² atoms

ethanol?

First need formula = C₂H₅OH

so weight = (2 [12.01]g/mole + 6 x [1.008 g/mole] + 16.00 g/mole) = 46.08 g/mole

then find moles = (1.00 g)/(46.08 g/mole) = 2.170 x10^-2mole

then multiply by Avogadro's Number: (2.170 x10^-2mole) x (6.022 x 10^-3 atoms/mole) = 1.307 x10²² ethanol molecules

But - there are 9 atoms/formula, So 1.176 x10²³ atoms = 1.18 x10²³ atoms

What is the mass of 3.01 x 10²⁰ atoms of carbon in:
- amu?

Each carbon atom weighs 12.01 amu, thus for 3.01 x 10²⁰ atoms have

mass = (3.01 x 10²⁰ atoms)(12.01 amu/atom) = 3.62 x 10²¹amu

First we need to find how many moles of carbon are present using Avogadro's Number:

(3.01 x 10²⁰atoms)/(6.02 x 10²³atom/mole) = 5.00 x 10^-4mole.

Now we multiply by the molecular weight to get the number of grams;
(5.00 x 10^-4mole)x 12.01g/mole = 6.00 x 10^-3g

What is the mass of iron in 200.0 mmoles of iron(III) sulfate

First need to determine the formula in order to know how many iron atoms are in it:

Fe₂(SO₄)₃

So for every mole of iron(III) sulfate there are two moles of iron, so we have 2 x 200.0 mmoles = 400.0 mmoles of Fe

Finally, (0.4000 mole Fe) x (55.85 g/mole Fe) = 22.34 g