### Richard A. Paselk

Chem 107

Fundamentals of Chemistry

Fall 2009

Lecture Notes: Concentration Examples

Worked Examples

% 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

• g?

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

• Making molar solutions
• Make up a 1.000 L solution of 0.25 M NaCl (note that water is the "default" solvent).

First weigh out 0.25 moles of NaCl

= (0.25 mole)(22.99 g + 35.45 g)/mole = 14.6 g

• What is the concentration of a solution made by dissolving 10.00 g of KI in enough water to make 1.000 L

First need to find the number of moles of KI:

(10.00 g) / ({39.10 g+ 126.9 g}/mole) = 6.024 x 10-2 mole

Thus the concentration will be 6.024 x 10-2 M

• Dilution problems
• What is the concentration of a solution resulting from adding 25.0 mL of 0.60 M CaCl2 to 475 mL of water.

First convert volumes to Liters: 25.0 mL = 0.0250 L; 475 mL = 0.475 L

Can solve as a ratio: (0.0250 L) (0.60 M) = (0.475L+ 0.025L) (x)

x = (0.60 M)(0.0250 L)/(0.500 L) = 3.000 x 10 -2

= 3.0 x 10 -2 M

• How much 1.000 M MgSO4 is needed to make 500.0 mL of a 0.25 M solution.

First convert volumes to Liters: 500.0mL = 0.5000 L

Can solve as a ratio: (0.5000 L) (0.25 M) = (1.0000 M) (x)

x = (0.25M)(0.5000 L)/(1.0000 M) = 0.125 L

= 0.13 L

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