NH3 ↝ always 1 N & 3 H ↝ ∴ in 17 g of ammonia, always 14 g N & 3 g
H
H2O ↝
always 2 H & 1 O ↝ ∴ in 18 g of water, always 2 g H & 16 g O
Percent Composition
Answer:
assume we have exactly 1 mole of C6H12O6
and thus,
m = nM = 1 mol(180.16 g/mol) = 180.16
g
C ↝ m = nM = 6 mol(12.011 g/mol) = 72.066 g
H ↝ m = nM = 12 mol(1.008 g/mol) = 12.10 g ↝ 180.16 g
O ↝ m = nM = 6 mol(16.00 g/mol) = 96.00 g
%C = (72.066g/180.16g) x 100 = 40.000%
%H = (12.10g/180.16g) x 100 = 6.716% ↝
100 %
%O = (96.00g/180.16g) x 100 = 53.29%
- We must always make this assumption as part of the solution, so that we can calculate the mass of the compound.
- If we have exactly 1 mol of C6H12O6, then we have exactly 6 mol C, 12 mol H and 6 mol O because of the ratio of elements found in the compound formula.
- To determine the mass of the individual elements, multiply the number of moles of each by their respective molar masses (all these masses should add up to the total mass of the compound).
- To determine the percentage composition for each element: take the mass of each element and divide it by the mass of the whole compound; multiply by 100 to make it a percentage rather than a decimal (all these percentages should add up to 100%).
↝ here are a couple of helpful videos (video 1 and video 2), just be sure to use the format that I've shown in the example above.
Empirical & Molecular Formulae
We can use percentage composition values to calculate the empirical or molecular formula of a compound.
The empirical formula (EF) is the lowest ratio of atoms in the compound and the molecular formula (MF) is the exact number of atoms in a compound
For example,
benzene: C6H6
(MF) and CH (EF)
dintrogen tetroxide: N2O4
(MF) and NO2 (EF)
magnesium chloride: MgCl2 (EF) and MgCl2 (MF)
ex. The percentage composition of
a compound is found to be 38.71% carbon, 9.71% hydrogen and 51.58 %
oxygen. (a) Determine the empirical
formula and (b) given that the molar mass of the unknown is 93.12 g/mol, find the molecular
formula of the compound.
Answer:
Assume we have exactly 100 g sample of the unknown compound.
C ↝ m = 38.71 g
n = m/M = 38.71 g/12.011 g/mol = 3.22 mol 3.22/3.22 = 1.00
n = m/M = 38.71 g/12.011 g/mol = 3.22 mol 3.22/3.22 = 1.00
H ↝ m = 9.71 g
n = m/M = 9.71 g/1.008 g/mol = 9.63 mol 9.63/3.22 = 2.99
n = m/M = 9.71 g/1.008 g/mol = 9.63 mol 9.63/3.22 = 2.99
O ↝ m = 51.58 g
n = m/M = 51.58 g/16.00 g/mol = 3.22 mol 3.22/3.22 = 1.00
n = m/M = 51.58 g/16.00 g/mol = 3.22 mol 3.22/3.22 = 1.00
empirical formula ↝ CH3O,
which has a molar mass
of (12.01g/mol + (3×1.008g/mol) + 16.00g/mol) = 31.04g/mol
93.12 g/mol/31.04 g/mol = 3.000, so 3(CH3O) = C3H9O3 which is the molecular formula
- We must make this assumption (exactly 100 g) as part of the solution so that when we use the given percentages in the calculations, it is simpler to do the math.
- We are given a percentage composition for each element. This is used to determine the mass of each element (for instance, 38.71% of the assumed 100 g sample is made up of C, so we have 38.71 g of C).
- We determine the number of moles of each element in the 100 g sample by using their molar masses.
- Now, we must determine the simplest whole number ratio of each element in the compound, using the number of moles of each. Sometimes, it is difficult to figure out the ratio just by looking at the number of moles of each (3.22:9.63:3.22 ???). So, here's a tip: find the lowest number of moles and divide all the moles by that number (for instance, 3.22 is the lowest number of moles, so each value is divided by 3.22 ↝ 3.22/3.22 = 1.00, 9.63/3.22 = 2.99, 3.33/3.22 = 1.00).
- Since 2.98 is very, very close to 3.00, we round that value to 3.00. This gives us the ratio of C:H:O of 1:3:1 and we use this simplest ratio to give us the empirical formula.
- To determine the molecular formula, we must find out how many empirical formulae fit into the molecular formula. So, we need to determine the molar mass for the empirical formula.
- Divide the molar mass of the molecular formula (given in the question) by the molar mass of the empirical formula (that we just calculated). This tells us how many empirical formula units make up the molecular formula. In this case, it is 3, so the molecular formula is C3H9O3.
↝ here are a couple of helpful videos (empirical formula, molecular formula) just be sure to use the format that I've shown in the example above.
Homework (SCH 3U):
Mole Review Assignment - work on this to prep for the test. The entire page must be completed to be fully prepared for the test.Homework (SCH 4C): do #17-19
Success Criteria:
- you must use nomenclature skills to determine the formulae given names.
- you MUST use proper technique and format for calculations, as laid out here. Failure to use proper format will result in many lost marks.
- definitions (mole, Avogadro's number, molar mass, empirical formula, molecular formula, etc.)
- calculations to determine molar mass, number of particles (molecules, atoms), number of moles, mass, percentage composition by mass, empirical/molecular formulae
Student Questions:1. Hey miss, how do we deal with the molar mass for a hydrate?When calculating the molar mass of a hydrate, the water is included in the calculation. For instance, to calculate the molar mass of CuSO4·5H2O, we would have 1 Cu, 1 S, 9 O and 10 H all added up together.
2. In a multi-step calculation, do I round as I go or wait 'til the end?
Use ALL the digits for every step in the calculation. Only round the final answer.
I will often appear to be rounding answers as I go through a multi-step calculation (mainly because I'm lazy and don't want to write down a bunch of digits), but I am ALWAYS actually using ALL the digits for the subsequent steps in the solution, until I get to the very last step. At that point, I round.
