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Chapter 9

Chemical Calculations: The Mole concept and Chemical Formula This material is not included in Midterm 1

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Law of Definite Proportions (John Dalton) Chapter 9

A given compound always contains the same proportion, by mass, of the elements.

Decompose samples of a pure compound to give the constituent elements: NO2

Sample Mass

decompose →

Mass of nitrogen

Mass of oxygen

A 5.362 g 1.632 g 3.730 g

B 10.613 g 3.230 g 7.383 g

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Percent Composition

What is the percent of nitrogen and percent of oxygen contained in each sample of NO2? Sample A: Sample B:

oxygen % 69.56 = % 100 x g 5.362g 3.730

nitrogen % 30.44 = % 100 x g 3625g 6321

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Formula Unit and Formula Mass

  Formula Unit   gives proportions of elements in a pure compound

NO2 CaCl2 molecular ionic

  Formula Mass   sum of atomic masses of atoms in one formula unit   Atomic masses are obtained from the periodic table in amu

(atomic mass units) eg: NO2 atomic mass of N 14.01amu

atomic mass of O 16.00 amu formula mass of NO2 : 14.01 + 2(16.00) amu = 46.01 amu

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Percent Composition from Formula

  NO2 formula mass 46.01 amu

total mass from each element x 100 %

formula mass

% N 14.01amu / 46.01 amu x 100% = 30.45 %

% O 2(16.00) amu / 46.01 amu x 100% = 69.55 %

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Percent Composition

Sample A Sample B

oxygen % 69.57 = % 100 x g 613.10g 7.383

nitrogen % 30.43 = % 100 x 613.10g 3.230

oxygen % 69.56 = % 100 x g 5.362g 3.730

nitrogen % 30.44 = % 100 x g 362.5g 632.1

g

30.45 %

69.55 %

From decomposition data From formula

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What is the formula mass of H2O?

  sum of the atomic masses of the total number of atoms represented by the formula

  2 x H = 2 (1.008) = 2.016 amu 1 x O = 1 x 15.994 = 15.994 amu

formula mass 18.010 amu = 18.010 amu

What is the formula mass of Ca(NO3)2?

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Counting Atoms, Ions and Molecules

Consider a reaction to form CO where: 1 atom of C combines with 1 atom of O to give CO, 1 molecule of

carbon monoxide 1 pair of C atoms + 1 pair of O atoms gives 1 pair of CO molecules 1 dozen C atoms + 1 dozen O atoms gives 1 dozen CO molecules 1 gross of C atoms + 1 gross of O atoms gives 1 gross CO molecules 1 mole of C atoms + 1 mole of O atoms gives 1 mole of CO molecules

1 mole = 6.022137 x 1023 objects Avogadro’s number

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One Mole

  If one mole of pennies was distributed equally to everyone in the earth’s population (~6 billion people), each person would have over a trillion dollars

  It would take 6 trillion galaxies the size of the Milky Way to provide 6.02 x 1023 stars

Mole Day   6:02 AM to 6:02 pm on October 23

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The Mole

 

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Converting between Moles and number of particles How many atoms are in 1.50 moles of Carbon?

1 mole = 6.02 x 1023 atoms

1.50 moles x 6.02 x 1023 atoms /mole = 9.03 x 1023 atoms How many molecules are in 5.35 moles of CO?

5.35 moles of CO x 6.02 x 1023 molecules/mole

Moles of substance Particles of a substance

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What is the Mass of one Mole ?

  Molar Mass of an Element The mass in grams that is numerically equivalent to the atomic mass of that element.

units of molar mass are: grams/mole atomic mass of 12C: 12 amu 1 mole of 12C atoms weighs 12 g/mole atomic mass of C : 12.011 amu 1 mole of C weighs 12.011 g/mole

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Molar Mass of a Compound

The mass in grams that is numerically equivalent to the formula mass of that compound.

NO2 formula mass: 46.01 amu

molar mass: 46.01 g/mole

The molar mass of any substance is the mass in grams of one mole of that substance.

units of g/mole

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What is the mass in grams of 0.250 moles of Na2CO3, an industrial chemical used in the making of glass? To go from moles to grams use molar mass as a conversion factor:

Molar mass of Na2CO3

2 x Na 2 (23.00) g/mole 1 x C 12.01 3 x O 3 (16.00) 106.01 g/mole

O.250 moles x 106.01 g/mole = 26.5 g of Na2CO3

How many moles of NH4Cl are in a 2.75 g sample of NH4Cl

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Substance Dissolved in a Liquid - concentration

  Molarity (Chapter 13.8 pg 471-477)

The number of moles of substance (solute) dissolved in one litre of solution - gives us the concentration

Molarity (M) = moles of solute litres of solution

1 Molar solution of CuSO4 contains 1 mole of CuSO4 in enough water to make up 1 litre of solution

0.5 Molar solution contains 0.5 moles of CuSO4 in 1 litre of solution

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Molarity in Calculations

2.5 moles of NaCl are dissolved in enough water to make 3.0 L of solution. What is the concentration, in molarity, of the solution.

5.5 g of KCl are dissolved in water to make 350 mL of solution. What is the molarity of the final solution?

2.5 moles of NaCl 3.0 L of solution

= 0.83moles/litre = 0.83 molar = 0.83M

5.5 g KCl = 0.07377 moles KCl 74.55 g/mole 0.07377 moles KCl = 0.21077 moles/L = 0.21 moles/litre 0.350 L

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Calculations Involving Moles

Moles of substance Particles of a substance

Grams of substance Moles of substance

Volume of a solution Moles of substance

Avogadro’s number

molar mass

molarity

particles/mole

g/mole

moles/L

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Example

How many moles of sulfuric acid, H2SO4, are contained in 0.80 L of a 0.050 M solution of the acid? volume = 0.80 L molarity, M = 0.050 moles/L # moles = 0.80 L x 0.050 moles

L = 0.040 moles (2 sig figs allowed)

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Chemical Formula of Compounds

  Formula give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions).

1 molecule of SO2 1 mole of SO2:

  If we know, or can determine, the relative number of moles

of each element in a compound we can determine a formula for the compound.

1 atom of S for every 2 atoms of O

1 mole S atoms for every 2 moles of O atoms

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Chemical Formula of Compounds

  Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms presents.

Ionic formula are always empirical formula

  Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

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Calculating Empirical Formula

  From experimental data: composition

• mass of individual elements • % composition of individual elements

  eg: decomposition of a known mass of sample to determine grams of each element.

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Formula

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34 g O. Determine a formula for this substance. We require mole ratios so must convert grams to moles moles of N = 2.34g of N = 0.167 moles of N

14.01 g/mole moles of O = 5.34 g = 0.334 moles of O

16.00 g/mole

Formula: 0.334 0.167ON 0.167 0.334 20.167 0.167

N O NO=

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Example:

A sample of a liquid compound contains 72.06 g of C and 6.06g of H. What is the chemical formula of the compound?

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Decomposition by Combustion

  compounds containing only carbon and hydrogen burn in oxygen to produce CO2 and H2O

•  calculate the mass of C from mass of CO2

•  calculate the mass of H from mass of H2O

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Sample Calculation

A sample of a pure substance containing only carbon and hydrogen weighs 1.025 g. When burned in oxygen, 3.007 g of CO2 and 1.845 g of H2O were produced. What is the empirical formula of the sample compound? moles of C in the compound from moles of CO2 (mass / molar mass) 3.007 g CO2 = 0.0683 mol of CO2 44.01 g/mol of CO2

0.0683 mol CO2 x 1 mol C = 0.0683 mol of C 1 mol CO2

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moles of H in the compound from moles of H2O moles of H2O = mass of H2O

molar mass of H2O

= 1.845 g H2O = 0.1025 mol H2O 18.00 g /mol H2O

moles of H = moles of H2O x 2 mol H

mol H2O = 0.2050 moles of H

Formula based on relative number of moles: C0.0683H0.2050

  Simplest whole number ratio CH3 Empirical Formula

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Combustion Products:

  The following compounds are produced from each of the listed elements when samples containing those elements are burned in oxygen:

C gives CO2

H gives H2O N gives NO or NO2 S gives SO2

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Calculating Empirical formula

The combustion of 1.621 g of a substance that contains only C, H, and O results in the recovery of 3.095 g of CO2 and 1.902 g of H2O. What is the empirical formula of this substance?

-determine the moles of C based on moles of CO2 formed -determine the moles of H based on moles of H2O formed moles of O: -oxygen in the CO2 and the H2O comes from both the O in the sample and the O in the air

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Determining amount of oxygen from combustion data

Mass of sample = mass of O + mass of C + mass of H from moles of C calculate mass of C in the sample from moles of H calculate mass of H in the sample mass of O = mass of sample – mass of (C+ H) -determine the moles of oxygen from mass of oxygen -determine the empirical formula knowing moles of C, H, O

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Empirical Formula from % Composition

A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams:

This will contain 60.80 grams of sodium, 28.60 grams of B and 10.60 grams H. Determine the number of moles of each Determine the simplest whole number ratio

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To obtain an Empirical Formula

1. Determine the mass in grams of each element present, if necessary.

2. Calculate the number of moles of each element.

3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.

4.  If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers

* Be careful! Do not round off numbers prematurely

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Calculation of the Molecular Formula

molecular formula = n(empirical formula)

molar mass = n(empirical formula mass)

A compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol n = molar mass = 92.0 g/mol emp. f. mass 46.01 g/mol n = 2

2(NO2) = N2O4

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Example

An unknown compound has the empirical formula CH2O A sample of 0.25 moles weighs 45.03 grams. What is the molecular formula of this compound?

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