empirical formulas & percent composition
DESCRIPTION
Empirical Formulas & Percent Composition. CO 2 Carbon dioxide. BCl 3 boron trichloride. Molar Mass. How many moles of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O?. Molar mass of C 2 H 6 O = 46.08 g/mol Calc. moles of alcohol. - PowerPoint PPT PresentationTRANSCRIPT
Empirical Formulas & Empirical Formulas & Percent CompositionPercent Composition
CO2 Carbon dioxide
BCl3 boron trichloride
Molar MassMolar Mass
How many How many molesmoles of alcohol of alcohol are there in a “standard” can are there in a “standard” can of beer if there are 21.3 g of of beer if there are 21.3 g of CC22HH66O?O?
(a)Molar mass of C2H6O = 46.08 g/mol
(b)Calc. moles of alcohol
21.3 g • 1 mol
46.08 g = 0.462 mol
How many molecules of alcohol are there in a “standard” can of beer if there are 21.3 g of C2H6O?
= 2.78 x 1023 molecules
We know there are 0.462 mol of C2H6O.
0.462 mol • 6.022 x 1023 molecules
1 mol
How many atoms of C are there in a “standard” can of beer if there are 21.3 g of C2H6O?
= 5.57 x 1023 C atoms
There are 2.78 x 1023 molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
2.78 x 1023 molecules • 2 C atoms1 molecule
Empirical & Molecular Empirical & Molecular FormulasFormulas
A pure compound always consists of the same elements combined in the same proportions by weight.
Therefore, we can express molecular composition as PERCENT BY WEIGHT
Ethanol, C2H6O52.13% C13.15% H 34.72% O
Percent CompositionPercent CompositionPercent CompositionPercent CompositionConsider some of the family of nitrogen-Consider some of the family of nitrogen-
oxygen compounds:oxygen compounds:
NONO22, nitrogen dioxide, nitrogen dioxide
Structure of NOStructure of NO22
What is the weight percent of N and of O?What is the weight percent of N and of O?
Percent Percent CompositionComposition
Percent Percent CompositionComposition
Consider NOConsider NO22, Molar mass =, Molar mass =
Percent of N and of O?Percent of N and of O?
Wt. % O 2 (16 .0 g O per mole )46 .0 g
x 100 % 69 .6%Wt. % O 2 (16 .0 g O per mole )46 .0 g
x 100 % 69 .6%
Wt. % N = 14.0 g N
46.0 g NO2 • 100% = 30.4 %Wt. % N =
14.0 g N46.0 g NO2
• 100% = 30.4 %
What are percentages of N and O in NO?What are percentages of N and O in NO?
46 g/ mol46 g/ mol
Determining Determining FormulasFormulas
Determining Determining FormulasFormulas
In chemical analysis we determine the % by weight of each element in a given
amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.
PROBLEM: What is its empirical formula of a compound of B and H
if 81.10% is boron?
%BB + %H = 100%
100% - 81.10% B = 18.90% H.If there were 100.0 g of the compound:
there are 81.10 g of B and 18.90 g of H.
Calculate the number of MOLES of each.
Determination of empirical Determination of empirical formulas:formulas:
mass of eachmass of each
Determination of empirical Determination of empirical formulas:formulas:
mass of eachmass of each
Calculate the number of moles of each element assuming 100.0 g total.
81.10 g B • 1 mol
10.81 g = 7.502 mol B81.10 g B •
1 mol10.81 g
= 7.502 mol B
18.90 g H • 1 mol
1.008 g = 18.75 mol H18.90 g H •
1 mol1.008 g
= 18.75 mol H
Determination of empirical Determination of empirical formulas:formulas:
moles of eachmoles of each
Determination of empirical Determination of empirical formulas:formulas:
moles of eachmoles of each
Take the ratio of moles of B and H.
atoms combine in small whole number ratios
Determination of empirical Determination of empirical formulas:formulas:
moles ratiomoles ratio
Determination of empirical Determination of empirical formulas:formulas:
moles ratiomoles ratio
Always divide by the smaller number.
18.75 mol H and 7.502 mol B
7.502 mol 7.502 mol
2.5 mol H to 1.0 mol B
But we need a whole number ratio. .
2.5 mol H or 5 mol H
1.0 mol B 2 mol B
EMPIRICAL FORMULA = B2H5
If the empirical formula is B2H5,
what is its molecular formula?We need to do an
EXPERIMENT to find the MOLAR MASS.
MS gives 53.3 g/molCompare empirical mass
= 26.66 g/unit Find the ratio of these masses.
Molecular Molecular formula = formula = BB44HH1010
2 =
26.66
3.53 2 =
26.66
3.53
Determine the formula of a compound of Sn and I
using the following data.
Determine the formula of a compound of Sn and I
using the following data.
• Reaction of Sn and I2 is done using excess Sn.• Mass of Sn in the beginning = 1.056 g
• Mass of Sn remaining
= 0.601 g
• Mass of iodine (I2) used
= 1.947 g
Find the mass of Sn that combined with 1.947 g I2.
Mass of Sn initially = 1.056 g
Mass of Sn recovered = 0.601 g
Mass of Sn used = 0.455 g
Find moles of Sn used:
0.455 g Sn • 1 mol
118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •
1 mol118.7 g
= 3.83 x 10-3 mol Sn
Tin and Iodine Tin and Iodine CompoundCompound
Tin and Iodine Tin and Iodine CompoundCompound
Now find the number of moles of I2 that combined with 3.83 x 10-3 mol Sn. Mass of I2 used was 1.947 g.
1.947 g I2 • 1 mol
253.81 g = 7.671 x 10-3 mol I21.947 g I2 •
1 mol253.81 g
= 7.671 x 10-3 mol I2
How many mol of iodine atoms? ?
= 1.534 x 10-2 mol I atoms= 1.534 x 10-2 mol I atoms
7.671 x 10-3 mol I2 2 mol I atoms
1 mol I2
7.671 x 10-3 mol I2 2 mol I atoms
1 mol I2
Tin and Iodine Tin and Iodine CompoundCompound
Now find the ratio of number of moles of moles of I and Sn that combined.
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn =
4.01 mol I1.00 mol Sn
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn =
4.01 mol I1.00 mol Sn
Empirical formula is Empirical formula is
SnISnI44
Practice Formula Determination
Homework:Worksheets pgs 17-20
Pg 13-14 good review for mole calculations
Prelab questions