MOLAR MASS &PERCENT COMPOSITION
MOLES
• How many moles are present?
CH4
C = 1 mol, H = 4 mol.
MOLAR MASS
CH4
Mass of 1 mol of C = 1 x 12.0 g = 12.0 g
Mass of 4 mol of H = 4 x 1.0 g = 4.0 g
Mass of 1 mol of CH4 = 16.0 g
Molar mass of any substance is the mass (in grams) of 1 mol of the substance. The molar mass is obtained by summing the masses of the component atoms.
MOLAR MASS
H2O
Mass of 2 mol of H = 2 x 1.0 g = 2.0 g
Mass of 1 mol of O = 1 x 16.0 g = 16.0 g
Mass of 1 mol of H2O = 18.0 g
PERCENT COMPOSITION
Mass percent = mass element present x 100
mass compound
The % mass that one particular element contributes to an The % mass that one particular element contributes to an entire compound.entire compound.
• to calculate… to calculate…
~~ find total mass of one molecule find total mass of one molecule
~~ ÷÷ mass mass ofof element element (part)(part) byby total mass (whole)total mass (whole)
~ x by 100 ~ x by 100
EXAMPLE• What is the Mass Percent of Na2S?
• Step 1 find Molar Mass.
• Na = 2 x 23.0 = 46.0 g
• S = 1 x 32.0 = 32.0 g
• Molar Mass = 78.0g
• Step 2 divide individual elements mass by the molar mass of entire compound.
Sodium = (46.0 g / 78.0 g) x 100 = 59.0%
Sulfur = (32.0g / 78.0g) x 100 = 41.0%
PRACTICE #1
• Calculate the mass percent of C10H14O.
• Carbon = 10 x 12.0g = 120.0g
• Hydrogen = 14 x 1.0g = 14.0g
• Oxygen = 1 x 16.0g = 16.0g
• Molar mass = 150.0g
PRACTICE #1
Mass Carbon = 120.0g /150.0g x 100 = 80.0%
Mass Hydrogen = 14.0g/150.0g x 100 = 9.3%
Mass Oxygen = 16.0g / 150.0 g x 100 = 10.7%
PRACTICE #2
• Determine mass percent of sulfuric acid.
• H4(SO4)2
• Hydrogen = 4 x 1.0g = 4.0g
• Sulfur = 2 x 32.1g = 64.2g
• Oxygen = 8 x 16.0g = 128.0g
• Molar Mass = 196.2g
PRACTICE #2
Mass Hydrogen = 4.0g / 196.2g x 100 = 2.04%
Mass Sulfur = 64.2g / 196.2g x 100 = 32.7%
Mass Oxygen = 128.0g / 196.2g x 100 = 65.2%
PRACTICE #3
• Determine mass percent of sulfuric acid.
• Ca(NO3)2
• Calcium = 1 x 40.1g = 40.1g
• Nitrogen = 2 x 14.0g = 28.0g
• Oxygen = 6 x 16.0g = 96.0g
• Molar Mass = 164.1g
PRACTICE #3
Mass Calcium = 40.1g / 164.1g x 100 = 24.4%
Mass Nitrogen = 28.0g / 164.1g x 100 = 17.1%
Mass Oxygen = 96.0g / 164.1g x 100 = 58.5%
EMPIRICAL AND MOLECULAR FORMULAS
• Empirical formula---lowest ratio of elements in compound
• Molecular formula ---actual ratio of elements in compounds
• Subscripts are mole ratios
• Compounds may have the same empirical formula but different molecular formulas.
• For example
CH is the empirical formula for C2H2 and C6H6
• If the subscripts can be reduced, then the formula is a molecular formula.
Empirical FormulasEmpirical Formulas
• The lowest whole # ratio of aThe lowest whole # ratio of acompound. (reduce the subscripts down if you can)compound. (reduce the subscripts down if you can)
• Ex. the empirical formula for glucose (CEx. the empirical formula for glucose (C66HH1212OO66) would ) would
be…be… CHCH22OO
• to calculate from % composition…to calculate from % composition…
1. 1. ÷÷ given % by atomic mass given % by atomic mass (to find moles)(to find moles)
2.2. ÷÷ each mole in step 1 by lowest mole value each mole in step 1 by lowest mole value (to find ratio) (to find ratio)
3. 3. values from step 2 = subscripts values from step 2 = subscripts
~~ subscripts subscripts MUSTMUST be a whole # be a whole # (multiply if they are not!)(multiply if they are not!)
Empirical FormulasEmpirical Formulas
• practice…practice…
1.1. What is the empirical formula of a compound thatWhat is the empirical formula of a compound that
contains 36.5% sodium, 25.4% sulfur and 38.1% oxygen?contains 36.5% sodium, 25.4% sulfur and 38.1% oxygen?
NaNa == 36.5%36.5%
23g23g== 1.59 moles1.59 moles
SS == 25.4%25.4% 32g32g
== 0.794 moles0.794 moles
OO == 38.1%38.1%
16g16g== 2.38 moles2.38 moles
******
0.7940.794== 22
0.7940.794 == 11
0.7940.794== 33
subscriptssubscripts
Na SONa SO22 33
Empirical FormulasEmpirical Formulas
• more practice…more practice…
2.2. What is the empirical formula of a compound thatWhat is the empirical formula of a compound that
contains 72.4% iron (Fe) and 27.6% oxygen?contains 72.4% iron (Fe) and 27.6% oxygen?
FeFe == 72.4%72.4%
5656== 1.293 moles1.293 moles
OO == 27.6%27.6%
1616== 1.725 moles1.725 moles
****** 1.2931.293
== 11
1.2931.293== 1.331.33
MUST be a whole #!!!MUST be a whole #!!!
x 3 x 3 == 44
x 3 x 3 == 33
Fe OFe O33 44
FINDING MOLECULAR FORMULAFROM EMPIRICAL FORMULA
• You must be given the actual molecular molar mass of the compound in the problem.
• Divide the actual molecular mass by the empirical formula mass.
• Then multiply all the subscripts by your answer.
• For example the actual molar mass of the previous compound is 88.0g.
• The empirical formula mass is 44.0g
• 88.0g/44.0g = 2
C2H4O X 2 = C4H8O2
LEARNING CHECK!!!WHICH OF THE FOLLOWING ARE EMPIRICAL FORMULAS AND WHICH ARE MOLECULAR?• H2SO4
• C6H12O6
• CH2O
• NaCl
• C6H12
• P4O10
• empirical
• molecular
• empirical
• empirical
• molecular
• molecular
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