Unit 6The Mole: %

Composition and Emperical Formula

PLEASE GRAB A CALCULATOR FROM THE

BACKCHECK YOUR BOX

Molecular mass – the sum of the average atomic masses of all the atoms in a molecule

***Remember*** A molecule is a neutral group of atoms that are held together by covalent bonds!!!

Formula Mass

The Mole and Molar Mass Molar Mass

The atomic molar mass is the mass of one mole of atoms. This is the value found on your Periodic Table. All values are rounded to three decimal

places. The term molar mass is a general

term referring to the mass of one mole of a compound. Also called molecular mass, formula mass, molar mass. Add the molar mass together for each

atom bonded together

Molar Mass of Compounds The molar mass (mm) of a compound is determined the

same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2

Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride =

40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2

20

Ca  40.08 17

Cl

35.45

The Mole Atoms, ions and molecules are so small it is

impractical to refer to 1, 2 or 3 of these particles.

Mole: 6.02 x 1023 items. This is the number of atoms required to produce the atomic mass of an element when that mass is expressed in grams.

Example:1 atom Mg = 24.31 amu (4.04 x 10-23 g)

6.02 x 1023 atoms Mg = 24.31 g

Avogadro’s Number

Avogadro’s Number (symbol N) is the number of atoms in 12.01 grams of carbon.

Its numerical value is 6.02 × 1023.

Therefore, a 12.01 g sample of carbon contains 6.02 × 1023 carbon atoms.

The Mole Reading Balanced Equations

2 Mg(s) + O2(g) 2 MgO(s)

2 Mg atoms + 1 O2 molecule produces 2 MgO formula units.

12.04 x 1023 Mg atoms + 6.02 x 1023 O2 molecules produces 12.04 x 1023 MgO formula units.

2 mol Mg atoms + 1 mol O2 molecules produces 2 mol MgO formula units.

The terms atoms, molecules and formula units are usually omitted: Two moles magnesium react with one mole oxygen to produce two moles magnesium oxide.

The MoleOther Examples:

Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

One mole of copper

metal

reacts with two moles of silver

nitrate solution

to produce two moles of silver

metal

and one mole of copper(II)

nitrate solution.

How Big Is a Mole?

The volume occupied by one mole of softballs would be about the size of the Earth.

One mole of Olympic shot put balls has about the same mass as the Earth.

What is the molar mass of KCl?

What is the molar mass of Ba(NO3)2?

Molar Mass

What is the molar mass of CuSO4•5H2O?

Molar Mass

Calculations Involving Mass and Moles

Mass to Moles Calculations:

ma#

ssof m

molaoles

r =

massand n =

m

M

= × # of m1 mo

oll

? g x g

OR

Given number of grams

Calculated number of moles = x

Calculated Molar

1

Mass

Calculations Involving Mass and Moles

Example: 1000. g NaCl is how many moles?

M 58.44 g/mol = = = 17.11

m 10 mon l

00.

g NaCl

1 mol

58.44 gOR × 17.111 000 mol. g = NaCl

1 Na 1 x 22.99 = 22.991 Cl 1 x 35.45 = 35.45

58.44 g/mol

FlowchartAtoms or Molecules

Moles

Mass (grams)

Divide by 6.02 X 1023

Multiply by 6.02 X 1023

Multiply by atomic/molar mass from periodic table

Divide by atomic/molar mass from periodic table

molar mass Avogadro’s number Grams Moles particles

Everything must go through Moles!!!

Calculations

Calculations Involving Mass and Moles

Moles-to-Mass Calculations:

# of grams = moles x molar mass m = n M

Calculated molar mass

Given number of moles

Calculated number of grams = x

? mol# of = × x g

gra1

s m mol

OR

Calculations Involving Mass and Moles

Example: 0.21 mol NaCl is how many grams?

58.44 gOR

1 mo × = 10 2.21 g m

lol NaCl

m = n M = 0.21 mol x 58.44 g/mol = 12 g NaCl

1 Na 1 x 22.99 = 22.991 Cl 1 x 35.45 = 35.45

58.44 g/mol

Percent Composition Law of Definite Proportions:

mass of element X

mass omass pe

f comporce

und = ×nt X 100

Elements in a compound combine in a constant proportion by mass.

For example, 1 g hydrogen combines with 8 g oxygen to produce 9 g water.

If atoms of an element have a specific mass, it must be true that atoms combine in a specific ratio.

Commonly expressed as percent composition

Percent CompositionExample (using experimental data):0.250 g of magnesium are burned to produce 0.415 g of magnesium oxide. Determine the mass percent of magnesium in the compound.

0.250 g magnesium

0.415 g magn1

e0

si0 × = 6

um oxi0.2

de% Mg

Percent CompositionExample (using molar mass calculations):What is the percent by mass of each of the elements in magnesium sulfate, MgSO4?

1024

0 × = 2.31

120.0.19

38% Mg 1

30

2.0 × = 2

07

120.6.6

384% S

16

04.

0 × = 500

120.3.1

386% O

1 Mg 1 x 24.31 = 24.311 S 1 x 32.07 = 32.074 O 4 x 16.00 = 64.00

120.38 g/mol

4. Combustion analysis of 0.500 g of an unknown hydrocarbon yielded 1.541 g of CO2 and 0.710 g H2O.

a) What is the percent by mass of carbon in the compound?

2

2

12.01 g C

44.01 g CO1.541 g CO × 0.420= 5 g C

0.4205 g C

0.500 g hydroc × = 8

arb4.1

o100

n % C

Empirical/Molecular Formulas and REVIEW

EMPERICAL FORMULA

The Empirical Formula is the simplest integer ratio of the elements in a compound.

For example: C4H8 is NOT an empirical formula. CH2 IS.

EMPERICAL FORMULA

Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).

Consider the amounts you are given as being in units of grams.

Convert the grams to moles for each element.

Find the smallest whole number ratio of moles for each element.

EMPIRICAL FORMULA

EXAMPLE:

Find the empirical formula for a compound consisting of 63% Mn and 37% O

EMPIRICAL FORMULA

Assuming 100 g of the compound, there would be 63 g Mn and 37 g OLook up the number of grams per mole for each element using the Periodic Table. There are 54.94 grams in each mole of manganese and 16.00 grams in a mole of oxygen.63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O

EMPIRICAL FORMULA

Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. In this case there is less Mn than O, so divide by the number of moles of Mn:1.1 mol Mn/1.1 = 1 mol Mn2.3 mol O/1.1 = 2.1 mol OThe best ratio is Mn:O of 1:2 and the formula is MnO2The empirical formula is MnO2

Molecular Formula

Molecular Formula is the exact ratio of the elements in the compound.

You can find the Molecular Formula of a compound by using the empirical formula and the molar mass.

The mass of the molecular formula is called the molecular mass (in the same way that the mass of the empirical formula is called the formula mass)

Molecular Formula

1) Calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula in order to find the ratio between the molecular formula and the empirical formula.

2)Then multiply all the atoms by this ratio to find the molecular formula!

Molecular Formula

EXAMPLE:

What is the molecular formula of a compound that has a molecular weight of 240g? A 5.00g sample contains 2.00g of Carbon, 0.34g of Hydrogen and 2.69g of Oxygen.