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