Why is Knowledge of Composition Important?. Because everything in nature is either chemically or physically combined with other substances it is necessary to have knowledge of chemical composition. To know the amount of a material in a sample, you need to know what fraction of the sample it is.
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Because everything in nature is either chemically or physically combined with other substances it is necessary to have knowledge of chemical composition. To know the amount of a material in a sample, you need to know what fraction of the sample it is.
Calculating Atomic Mass
Gallium has two naturally occurring isotopes: Ga-69 with mass 68.9256 amu and a natural abundance of 60.11% and Ga-71 with mass 70.9247 amu and a natural abundance of 39.89%. Calculate the atomic mass of gallium.
Convert the percent natural abundance into decimal form.
Determine the Formula to Use
Atomic Mass = (abundance1)∙(mass 1) + (abundance2)∙(mass 2) + ...
Apply the Formula:
Atomic Mass = 0.6011 (68.9256 amu) + 0.3989 (70.9247 amu) = 69.72 amu
A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?
Given: 1.1 x 1022 Ag atoms
Wanted: ? Moles
1 mole Ag atoms = 6.022 x 1023 Ag atoms
= 1.8266 x 10-2 moles Ag
= 1.8 x 10-2 moles Ag
How do you calculate the number of moles of sulfur in 57.8 g of sulfur?
= 1.80 moles S
= 1.80287 moles S
How many aluminum atoms are in an aluminum can with a mass of 16.2 g?
= 3.6159 x 1023 atoms Al
Sig. Figs. & Round: = 3.62 x 1023 atoms Al
Calculate the mass (in grams) of 1.75 mol of water
= 31.535 g H2O
= 31.5 g H2O
Sig. Figs. & Round:
Find the mass of 4.78 x 1024 NO2 molecules?
= 365.21 g NO2
Sig. Figs. & Round: = 365 g NO2
L-Carvone, (C10H14O), is found in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquors, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.
Molar Mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O)
= 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mo
1 mole C10H14O = 150.2 g C10H14O
1 mole C10H14O 10 mol C
1 mole C = 12.01 g C
= 44.2979 g C
= 44.3 g C
Mass of Carbon = 2 moles C x (12.01 g) = 24.02 g
Mass of Hydrogen = 6 moles H x (1.008 g) = 6.048 g
Mass of Oxygen = 1 mol O x (16.00 g) = 16.00 g
1 mole C2H5OH = 46.07 g
Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g O and the rest H?
C = 24.0 g, O = 3.2 g
H = 30.0 g – (24.0 g + 3.2 g) = 2.8 g
Hydrates are compounds that include water molecules as part of their structure
Na2CO3 • 10 H2O
NiCl2 • 6 H2O
What does this have to do with percent composition?
Let’s say we take 1.20 g of NiCl2 • x H2O (a hydrate)
With lots of heat from a Bunsen burner, we remove all the water and are left with 0.65 g of NiCl2. What is the percent composition of water in the hydrate?
(1.20 g of NiCl2 • 6 H2O) - (0.65 g of NiCl2) =
= 0.55 g H2O
% Comp. H2O = 0.55 g H2O (part) x 100% = 45%
1.20 g NiCl2 • x H2O (whole)
*We didn’t need to know the exact composition of the NiCl2 hydrate
We find a compound and determine that it is composed of 31.2% nitrogen and 68.8% oxygen. What is the chemical formula of the compound?
For every 100. grams of material, there is 30.9 g N and 69.1 g O. We first convert to moles.
30.9 g N 1 mole N = 2.21 mole N
69.1 g O 1 mole O = 4.32 mole O
Molecular Formula = H2O2
Empirical Formula = HO
Molecular Formula = C6H6
Empirical Formula = CH
Molecular Formula = C6H12O6
Empirical Formula = CH2O
Examples: Empirical Formulas
A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Example:Find the empirical formula of aspirin with the given mass percent composition.
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
1 mole C = 12.01 g C
1 mole H = 1.01 g H
1 mole O = 16.00 g O
This gives the formula of
C4.996H4.44O2.220. But, the formula
must have the atoms in whole
Multiplying the subscripts by 4 gives the final ratio!
All these molecules have the same Empirical Formula. How are the molecules different?
Molar Massempirical formula
= factor used to multiply subscripts
Determine the Molecular Formula of Cadinene if it has a molar mass of 204 g/mol and an empirical formula of C5H8
Multiply the empirical formula by the factor above (3) to give the molecular formula: