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# Ch 7.3 Using Chemical Formulas - PowerPoint PPT Presentation

Ch 7.3 Using Chemical Formulas. The Mass of a Mole of an Element. Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu). Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol). Carbon = 12.01 g/mol Hydrogen = 1.01 g/mol

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### Ch 7.3 Using Chemical Formulas

• Remember: The atomic mass of an element (a single atom) is expressed in atomic mass units (amu).

• Molar Mass: is the atomic mass of an element expressed in grams/mole (g/mol).

• Carbon = 12.01 g/mol

• Hydrogen = 1.01 g/mol

• When dealing with molar mass, round off to two decimals. 12.011g/mol -> 12.01g/mol

• You calculate the mass of a molecule by adding up the molar masses of the atoms making up the molecules.

• Example: H2O

• H = 1.01 g x 2 atoms = 2.02 g/mol

• O = 16.00 g x 1 atom = 16.00 g/mol

• Molar Mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol

• This applies to both molecular and ionic compounds

• Find the molar mass of PCl3

• P = 30.97 g x 1 atom = 30.97 g/mol

• Cl = 35.45 g x 3 atoms = 106.35 g/mol

• PCl3 = 30.97 g + 106.35 g = 137.32 g/mol

• What is the molar mass of Sodium Hydrogen Carbonate (NaHCO3) ?

• Na = 22.99 g x 1 atom = 22.99 g/mol

• H = 1.01 g x 1 atom = 1.01 g/mol

• C = 12.01 g x 1 atom = 12.01 g/mol

• O = 16.00 g x 3 atoms = 48.00 g/mol

• NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00

= 84.01 g/mol

• You can use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance.

• Mass (g) = # of moles x mass (g)

1 mole

Example: If molar mass of NaCl is 58.44 g/mol, what is the mass of 3.00 mol NaCl?

Mass of NaCl = 3.00 mol x 58.44g =

1 mol

175 g NaCl

• What is the mass of 9.45 mol of Aluminum Oxide (Al2O3)?

• Find molar mass of Al2O3

= 101.96 g/mol

• Mass = 9.45 mol Al2O3 x 101.96 g Al2O3

1 mol Al2O3

= 964 g Al2O3

• You can invert the conversion factor to find moles when given the mass.

• Moles = mass (g) x 1 mole

mass (g)

Example: If molar mass of Na2SO4 142.05 g/mol, how many moles is 10.0 g of Na2SO4?

Moles of Na2SO4 = 10.0 g x 1 mol =

142.05 g

= 0.0704 mol Na2SO4

• How many moles are in 75.0 g of Dinitrogen Trioxide?

• Find molar mass of N2O3

= 76.02 g/mol

• Moles = 75.0 g N2O3 x 1 mole =

76.02 g

N2O3

0.987 mol N2O3

• Percent Composition: the relative amount of the elements in a compound.

• Also known as the percent by mass

• It can be calculated in two ways:

• Using Mass Data

• Using the Chemical Formula

% mass of element= mass of element x100% mass of compound

• When a 13.60 g sample of a compound containing Mg and O is decomposed, 5.40 g O is obtained. What is the % composition of this compound?

Mass of compound: 13.60 g

Mass of oxygen: 5.40 g O

Mass of magnesium: 13.60 g - 5.40 g = 8.20 g Mg

% Mg = 8.20 g Mg x 100% =

13.60 g

% O = 5.40 g O x 100% =

13.60 g

60.3%

39.7%

• Find the percent composition of Cu2S.

• Find mass of Cu and S

• Cu = 63.55 x 2 = 127.10 g

• S = 32.07 g

• Find mass of Cu2S

• 127.10 g + 32.07 g = 159.17 g

% Composition

• Cu = 127.10 g x 100% =

159.17 g

• S = 32.07 g x 100% =

159.17 g

79.85%

20.15%

• 7.3 pg 253 #30-33

### Ch 7.4Determining Chemical Formulas

• Empirical Formula: shows the smallest whole-number ratio of the atoms of the elements in a compound.

• Example:

• The Empirical Formula for Hydrogen Peroxide (H2O2) is HO with a 1:1 ratio.

• The Empirical Formula for Carbon Dioxide (CO2) is CO2 with a 1:2 ratio.

• A compound is found to contain 25.9% Nitrogen and 74.1% Oxygen. What is the Empirical Formula of the compound?

• 25.9 g N x 1 mol N = 14.01 g N

• 74.1 g O x 1 mol O = 16.00 g O

• N1.85O4.63 = N1O2.5 = N2O5

1.85 mol N

4.63 mol O

• Molecular Formula: tells the actual number of each kind of atom present in a molecule of a compound

• Example:

• The Molecular Formula for Hydrogen Peroxide is H2O2.

• The Molecular Formula for Carbon Dioxide is CO2

• It is possible to find the Molecular Formula using the Empirical Formula if you know the molar mass of the compound.

• Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N

• Step 1: Find the empirical formula molar mass

• 12.01 + (4 x 1.01) + 14.01 = 30.06 g/mol

• Step 2: Divide molar mass by EF molar mass

• 60.0 g/mol = 1.99  2 30.06 g/mol

• Step 3: Multiply empirical formula by 2

• CH4N x 2 = C2H8N2

• 7.4 pg 253 #36-38