2013/2014 1

STOICHIOMETRIC CALCULATIONS

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Concentration Expression

Basic units to describe the total amount of chemical species.

Example:

g Na2SO4 = moles x F wt. = moles x 142.04 g/mol 3

Cont…

Note: g/mol is the same as mg/mmol, g/L the same as mg/ml, and mol/L the same as mmol/mL.

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Periodic table

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Exercises..

Calculate the number of moles in 500 mg Na2WO4. (Ans:0.00170 mol)

How many grams are in 0.1 mol of NaOH ? (Ans:4g)

How many milligrams are in 0.250 mmol Fe2O3? (Ans:39.9 mg)

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2013/2014 2

Molarity

1 mole of substance in 1 L (dm3) of a solution

Millimoles (mmol) = Molarity x mL

Molarity (M) = Volume of solution (L)

Amount of solute (mol solute)

Volume of solution (ml)/ 103

Mass of solute (g) / F wt. (gmol-1) =

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Examples

1. How many moles of solute are present in 1.5dm3 of 0.70M sodium hypochlorite?

Ans: 1.05 mol NaClO

2. One saline solution contains 0.90g NaCl in exactly 100ml of solution. What is the molarity of the solution?

Ans: 0.15 M

3. You have 10.8 g potassium nitrate. How many mL of solution will make this a 1.4 M solution?

Ans: 76.3 ml

4. How many grams per milliliter of NaCl are contained in a 0.250 M solution?

Ans: 0.0146 g/ml 5. How many grams Na2SO4 should be weighed out to

prepare 500mL of a 0.100 M solution? Ans: 7.10 g

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Molality

The number of moles of solute per kilogram of solvent.

Molality (m) =

Kg of Solvent

mol of solute

1 kg water = 1 L water

mass of solvent only kg 1

mol0.25 0.25m

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Examples

1. Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.

Ans: 3.2 m MgCl2

2. How many grams of NaCl are required to make a 1.54m solution using 0.500 kg of water.

Ans: 45.0 g NaCl

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9. Normality

*** http://faculty.chemeketa.edu/lemme/CH%20122/handouts/normality.pdf

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Normality

Is the number of equivalent (eq) per unit volume .

Normality,

Equivalent weight (EW) is defined as the ratio of a chemical species formula weight (FW) to the number of its equivalents;

Equivalent weight,

An equivalent represents the mass of reacting units (proton or electron).

Number of equivalent of a substance (eq) from the number of gram can be calculated by:

Therefore the normality of the solution can be calculated by:

It is an older unit of concentration that is frequently ignored in today’s laboratory

eq

FWEW

mL

meq

L

eqN

)/()/( meqmgwteq

mgmeq

eqgwteq

geq

L

eqgwteqg

L

eqN

)/(/

mL

meqmgwteqmg

mL

meqN

)/(/

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2013/2014 3

Calculate the equivalent weights and normality of the following substances :

98.2g/L of KMnO4 (Mn(VII)) is reduced to Mn2+

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

EW = 158.04 g/mol = 31.608 g/eq 5 eq /mol Normality = 98.2 g/L = 3.1 N 31.608 g/eq

Example

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Concentrations of Solutions

For very dilute solutions

Part per million (ppm)

ppm = volume solution (L) mass solute (mg)

Part per billion (ppb)

ppb = volume solution (L)

mass solute (µg)

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Examples

2. What is concentration (in ppm) of K+ in 0.025 M K2SO4 solution?

ppm

L

K mg

g

mg

K 1mol

g

SO K mol

K mol

L

SO K mol

1955

1955 1000 10 . 39

1

2 02500 . 0

4 2

4 2

+

+

+

1. How many grams of Zinc are required to prepare a 125ml solution which the concentration is 1.25 ppb.

Ans: 0.156 mg

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5.1034 g KH2PO4 were dissolved in 250ml solution. Calculate: (K=39.1, H=1, P=31, O=16)

a) What is the molarity of the solution?

(Ans: 0.15 M)

b) ppm of PO4 of the solution? (Ans: 14250 ppm)

c) ppm of K of the solution? (Ans: 5865 ppm)

d. ppm of H of the solution? (Ans: 300 ppm)

e) ppm of P of the solution? (Ans: 4650 ppm)

Exercises..

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1. Weight/Weight Percent (w/w%)

2. Volume/Volume Percent (v/v%)

3. Weight/Volume Percent (w/v%)

These units express concentration as units of solute per 100 units of sample

A weight/weight percent is defined as:

A volume/volume percent is defined as:

A weight/volume percent is defined as:

Example: A solution in which a solute has a concentration of 23% w/v contains 23g of solute per 100 ml of solution.

100%/

solutionofgrams

soluteofgramsww

100%/

solutionofvolume

soluteofvolumevv

100%/

solutionofvolume

soluteofgramsvw

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Example

A sample weighing 1.2304 g contains 0.1012 g iron. Percentage of iron (w/w) = 0.1012 X 100 1.2304 = 8.22 %

OR

82.2 ppt iron, 82200 ppm iron

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2013/2014 4

Exercise

A sample weighing 1.3535 g contains 0.4701 g .Calculate the % Fe in the sample. What is the Fe content in ppt and ppm?

(Ans:Fe=34.73%; 347. 3 ppt; 3.473 x 105

ppm)

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Exercise

Calculate the molar concentrations of 1.00 ppm solution each of Li+ and Pb2+

(Ans :1.44x10-4mol/L Li; 4.83X10-6mol/L Pb) A 2.6 g sample of plant tissue was analyzed

and found to contain 3.6 μg zinc. What is the concentration of zinc in the Plant in ppm ? In ppb?

(Ans:1.4μg/g; 1.4x103ng/g) What weight of Pb(NO3)2 will have to be

dissolved in 1 liter of water to prepare a 100 ppm Pb 2+ solution?

(Ans: 0.137 g)

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Example

Calculate molar concentration of HNO3 (63.0g/mol) in a solution: specific gravity of 1.42 and is 70% HNO3 (w/w). Tips:

Given 70% (w/w) HNO3 70g HNO3/ 100 g solution

Need to convert to molarity of HNO3 Moles HNO3 per liter solution Specific Gravity g / mL

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Solution

g HNO3 = 1.42 g reag x 103 mL reag x 70 g HNO3

L reag mL reag 1 L reag 100g reag

= 994 g HNO3

L reagent

Mol HNO3 = 994 g HNO3 x 1 mol = 15.8 mol

L reagent L reagent 63 g HNO3 L

= 16 M

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Exercise

How many mL of 94.0% (g/100g) H2SO4 , density 1.831 g/ml, are required to prepare 1 L of 0.100 M solution? (FW: H2SO4 = 98.1g/mol)

(Ans: 5.7 mL)

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Dilution

The desired molarity solutions are often prepared from concentrated stock solutions by adding water.

M1 x V1 = M2 x V2

Moles of solute = moles of solute

before dilution after dilution

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2013/2014 5

Making a

Dilute

Solution

Concentrated

solution

remove

sample

Diluted solution

same number of

moles of solute

in a larger volume

mix

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Examples

1. How many mililiters of aqueous 2.00 M MgSO4 solution must be diluted with H2O to prepare 100 mL of aqueous 0.400 M MgSO4.

M1 = 2.00 M MgSO4 M2 = 0.400 M MgSO4 V1 = ?? V2 = 100 mL

V1 = 0.400 M X 100 mL = 20 mL MgSO4

2.00 M 26

28.9 M (0.075L) = 0.100 M (15.0 L)

2.1675 mol HAVE > 1.5 mol NEED

2. You have 75 mL of conc. HF (28.9 M); you need 15.0 L of 0.100 M HF. Do you have enough to do the experiment?

Cont…

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Exercise

You wish to prepare a calibration curve for the spectrophotometric determination of permanganate.You have a stock 0.100 M solution of KMnO4 and a series of 100 mL volumetric flasks. What volumes of the stock solution wll you have to pipet into flasks to prepare standards of 0.001, 0.002, 0.005, 0. 010 M KMnO4 solutions?

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Stoichiometric : relationship btwn the no. of moles of reactants & product as shown by a balanced equation.

Example :

2NaI + Pb (NO3) PbI2 + 2NaNO3

Can calculate the quantity of the reactant & product, limiting/excess reactants etc.

Definition

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2013/2014 6

1. Calculate how many moles O2 are required to completely react with 3.5g of H2 ?

O2 + 2H2 2 H2 O

2. Calculate how many grams of ammonia are

produced when you react 2.00g of nitrogen with excess hydrogen.

N2 + 3 H2 2 NH3

3.5 g H2 1 mol H2 1 mol 02

2.0 g H2 2 mol H2

= 0.875 mol O2

Stoichiometric Calculations

= 2.4g

2.00 g N2 1 mol N2 2 mol NH3 17.06g NH3

28.02 g N2 1 mol N2 1 mol NH3 31

Exercises

4NH3 + 5O2 6H2O + 4NO

a) How many moles of H2O are produced if 0.176 mol of O2 are used? (Ans: 0.2112 mol H2O)

b) How many grams of H2O are produced if 1.9

mol of NH3 are combined with excess oxygen? (Ans: 51.4 g H2O)

c) How many grams of NO can be made from 120

g of NH3? (Ans: 211 g NO)

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Limiting Reagents

Caution: this stuff is difficult to follow at first.

Be patient. 33

The reactant which runs out first = LIMITING REACTANT

Once it runs out, the reaction s.

That reactant is said to be in EXCESS (there is too much). Limiting Reactant Excess Reactant

•Used up in a reaction

•Determines the amount of product

•Added to ensure that the other reactant is completely used up

Cont…

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Example

2 NH3 + 3 CuO ---> N2 + 3 Cu + 3 H2O

18.1 g 90.4 g

Which reactant is limiting (LR), which is in excess (ER), and how much N2 is produced?

Solution:

CuO = (90.4 g /80 gmol-1) / 3 = 0.38 mol

NH3 = (18.1 g /17 gmol-1) / 2 = 0.53 mol

CuO = LR

NH3 = ER

90.4 g CuO 1 mol CuO 1 mol N2 28 g N2

80 g CuO 3 mol CuO 1 mol N2

= 10.55 g N2

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90.4 g CuO 1 mol CuO 2 mol NH3 17 g NH3

80 g CuO 3 mol CuO 1 mol NH3

= 12.81 g NH3

USED

Can we find the amount of excess NH3 ? How??

18.1 g NH3 – 12.81 g NH3 = 5.29 g NH3 (EXCESS)

Given amount

of excess

reactant

Amount of

excess

reactant

actually used

Note : started with the

limiting reactant! Once you

determine the LR, you

should only start with it!

Cont…

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2013/2014 7

Exercises

79.1 g of zinc react with 0.90 L of 2.5M HCl to produce hydrogen gas.

a) Write a balanced equation.

b) Identify the limiting and excess reactants.

c) How many liters of hydrogen are formed at STP?

(At STP : 1 mol gas = 22.4 dm3)

Ans: LR = HCl, ER = Zn

Ans: 25 L H2

Zn + 2HCl ZnCl2 + H2

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The theoretical yield of a chemical reaction is the calculated quantity of product in the reaction.

The actual yield is the amount you actually get when you carry out the reaction.

Actual yield will be LESS than the theoretical yield, for many reasons … can you name some?

Yields of Chemical Reactions

actual yield Percent yield = ––––––––––––– × 100 theoretical yield

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Example

When 45.8 g of K2CO3 react with excess

HCl, 46.3 g of KCl are formed. Calculate

the theoretical and % yields of KCl.

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g ? g

actual: 46.3 g

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45.8 g

K2CO3

1 mol

K2CO3

138.21 g

K2CO3

= 49.4g

KCl

2 mol

KCl

1 mol

K2CO3

74.55

g KCl

1 mol

KCl

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g ? g

actual: 46.3 g

Theoretical Yield

Cont…

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Theoretical Yield = 49.4 g KCl

% Yield = 46.3 g

49.4 g 100 = 93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g 49.4 g

actual: 46.3 g

Cont…

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2013/2014 8

Exercise

What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?

[Ans: % yield = 96.7%; 143 g (theoretical value)]

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