Ch 7 3 using chemical formulas
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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

Ch 7.3 Using Chemical Formulas


The mass of a mole of an element
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

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


The mass of a mole of a compound
The Mass of a Mole of a Compound

  • 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


Converting moles to mass
Converting Moles to Mass

  • 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


Example 2 moles to mass
Example 2: Moles to Mass

  • 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


Converting mass to moles
Converting Mass to Moles

  • 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


Example 2 mass to moles
Example 2: Mass to Moles

  • 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
Percent Composition

  • 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


Example
Example

  • 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%


Homework
Homework

  • 7.3 pg 253 #30-33


Ch 7 4 determining chemical formulas

Ch 7.4Determining Chemical Formulas


Empirical formulas
Empirical 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.


Determining the empirical formula of a compound
Determining the Empirical Formula of a Compound

  • 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 formulas
Molecular Formulas

  • 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.


Finding the molecular formula
Finding the Molecular Formula

  • 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


Homework1
Homework

  • 7.4 pg 253 #36-38


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