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Empirical & Molecular Formulas
Calculations
Solve This:A compound is put through a _______________ ______________ and found to have the following percent composition:
Ba: 69.58%
C: 6.10%
O: 24.32%
What is the empirical formula?
CombustionAnalyzer
How to start!Step 1:
Assume that you have a 100 gram sample
In a 100 gram sample you would have:
69.58g of Ba
6.10g of C
24.32g of O
Step 2 (mass to moles - m to n)
Calculate the number of moles of each represented element:
nBa: 69.58g / 137.33g/mol = 0.5067 mol
nC: 6.10g / 12.01g/mol = 0.508 mol
nO: 24.32g / 16.00g/mol = 1.520 mol
Step 3 (find the simplest whole number ratio)
Divide each mole value by the smallest mole value - Ba in this case (0.5067 mol):
Ba: 0.5067 / 0.5067 = 1.000 mol ~ 1 mol
C: 0.508 / 0.5067 = 1.00 mol ~ 1 mol
O: 1.520 / 0.5067 = 3.000 mol ~ 3 mol
So, our ratio is... Ba (1) : C (1) : O (3)
Therefore, our empirical formula is BaCO3
Practice: Q’s 1 - 6 (p. 186) & 1 - 6 (p. 189)
Question #6, pg 189
Propane is a hydrocarbon that is used as fuel in BBQs and some
cars. In a 26.80 g sample of propane, 4.90 g is hydrogen and the
remainder is carbon. What is the empirical formula of propane?
Step 1 (% composition)
Calculate the percent composition of hydrogen:
4.90g / 26.80g
= 18.3% hydrogen
Derive, from the above, the percent composition of carbon:
100% - 18.3%
= 81.7% carbon
Step 2 (assume 100g sample & find moles)
nH: 18.3g / 1.01g/mol
= 18.1 mol
nC: 81.7g / 12.01g/mol
= 6.80 mol
Step 3 (find the simplest whole number ratio)
Part a:H: 18.1 mol / 6.80 mol = 2.66 mol
C: 6.80 mol / 6.80 mol = 1 mol
Part b (we’re not done, because 2.66 is not whole!!!):
H: 2.66 mol x 3 = 7.98 mol ~ 8 mol H
C: 1 mol x 3 = 3 mol C
Therefore, the empirical formula is C3H8
McCloskey loves his __________Agents were worried about McCloskey’s weird behaviour, so they used a combustion analyzer to determine the percent composition of the compound believed to be causing the issue. The results were as follows:
C (49.5%), H (5.15%), N (28.9%), O (16.45%)
What is the empirical formula?
Step 1 (assume 100g sample & find moles)
nc: 49.5g / 12.01g/mol
= 4.12 mol
nH: 5.15g / 1.01g/mol
= 5.10 mol
nN: 28.9g / 14.01g/mol
= 2.06 mol
nO: 16.45 / 16.00g/mol
= 1.028 mol
Step 2 (find the simplest whole number ratio)
C: 4.12 mol / 1.028 mol = 4.01 ~ 4 mol
H: 5.10 mol / 1.028 mol = 4.96 ~ 5 mol
N: 2.06 mol / 1.028 mol = 2.00 ~ 2 mol
O: 1.028 mol / 1.028 mol = 1.000 ~ 1 mol
The empirical formula is therefore:
C4H5N2O
McCloskey loves his __________
You still don’t have enough information to figure out the compound causing McCloskey’s behavioural issues, but lucky enough, agents used a mass spectrometer to yield the molar mass:
It is 195.0 g/mol. What does McCloskey love?
Step 1 (calculate M for the empirical formula)
C4H5N2O= ! 4 x 12.01g/mol! ! ! ! ! 5 x 1.01g/mol! ! ! ! ! 2 x 14.01g/mol! ! ! ! ! 1 x 16.00g/mol
! ! ! ! ! 97.11g/mol
Ask yourself: “how many times will the mass of my empirical formula fit into the molar mass as determined by the mass spectrometer?”
Step 2 (divide M by the mass of the empirical formula
195.0g/mol / 97.11g/mol = 2.008 ~ 2 xTherefore, if your empirical formula is C4H5N2OThen your molecular formula is
C4H5N2O x 2 = C8H10N4O2
So, what does McCloskey love?
HomeworkLook closely @ Sample Problem #3 on p. 192 AND the “Summary” on p. 193.
Realize that if you are given both the % composition & the molar mass (M), you can determine the molecular formula in one step
Do Questions 8-10 on p. 193