Formula Mass
Also called Molar Mass The mass of a compound. ¨ Units are in g/mol: grams per mole ¨ Coefficients are the big numbers in front of the compound—how many compounds/molecules ¨ Subscripts are the small numbers to the bottom right of the symbol—how many atoms there are ¨ ¨ Molar Mass
Example 1
NaCl
¨ Mass of Na=23.00 g/mol
¨ Mass of Cl=35.45 g/mol
¨ 1 atom of Na
¨ 1 atom of Cl
¨
NaCl
¨ Total mass of Na = 1x 23.00 g/mol = 23.00g/mol ¨ Total mass of Cl = 1 x 35.45 g/mol = 35.45 g/mol ¨ Mass of compd. = 23.00 + 35.45 = 58.45 g/mol Percent Composition
The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100. ¨ Can be found 2 ways: ¨ Percent Composition
¤ Experimentally ¤ Calculate % from a given formula
Formula:
Example 1
Part/whole x 100
¨ Find the % composition of a compound that contains 2.30g Na, 1.60g O, and 0.100g H in a 4.00g sample of the compound. % mass of element X= grams of X • 100
grams of cmpd
Solution
¨
Determine % of each element
% Na = 2.30 = 0.575 x 100 = 57.5% Na
4.00
Example 2
¨ A sample of 2.45g of Aluminum oxide decomposes into 1.30g of Al and 1.15g of O. What is the percentage composition of the compound? % O = 1.60 = 0.4 x 100 = 40.0% O
4.00
% H = 0.100 = 0.025 x 100 = 2.50% H
4.00
**Make sure all percentages add up to 100% !
Solution
Example 3
% Al = 1.30 x 100 = 53.1% Al
2.45
¨ % O = 1.15 x 100 = 46.9% O
2.45
¨ An 8.20‐g piece of Mg combines completely with 5.40‐g of oxygen to form a compound. What is the percent comp. of this compound? Info: ¤ Mass of Mg=8.20 g ¤ Mass of O=5.40 g ¤ Mass of compound= 8.20+5.40g=13.60g ¤ % Mg =? ¤ % O=?
Example 4
¨
% mass of Mg= grams of Mg
grams of cmpd
x 100
¨ ¨ % Mg= 8.20g/13.60g x 100% = 60.3%
% O=5.40g/13.60g x 100% = 39.7%
Calculate the percent comp. of propane (C 3 H 8 ). Info: ¤ Molar Mass of propane=44.0 g/mol ¤ Mass of C in propane= 36.0 g ¤ Mass of H in propane=8.0 g ¤ Percent H=? ¤ Percent C=? Warm Up
¨
% mass of H= grams of H x 100
grams of cmpd
% H = (8.0g / 44.0g) x 100 = 18.2%
¨
Determine if the following are the same
compound.
¤ 45g sample containing 35.1g Fe & 9.9g O
¤ 215g sample containing 167.7g Fe & 47.3g O
% C=(36.0g / 44.0 g) x 100 = 81.8%
Empirical & Molecular Formulas
¨ Empirical and Molecular Formulas
¨ Empirical Formula ‐ a compound/molecule with the smallest whole‐number ratio of elements Molecular Formula – specifies the actual number of atoms of each element in a compound/molecule of a particular substance ¤ will be a multiple of an empirical formula
Example of Empirical
¨ 3. Suppose you are given a cmpd with a % composition by mass that is 80% C and 20% H. What is the empirical formula? Convert g of each element to moles of that element. ¤ Solution 1 1. 2. 4. If you are given gramsà go straight to moles! 80g C 1 mol C = 6.7 mol C 12.01g C Divide each # of moles by the smallest value to get a ratio. 6.7 mol C = 1 6.7 mol C 5. 20 mol H = 2.98 6.7 mol C Must be whole numbers! Round up to next whole number 6. 20g H Assume you have 100g of sample If you have 100g, then the mass of each element will have the same # as the %. 80% C = 80g C 20% H = 20g H 1 mol H = 20 mol H 1.01g H Write the formula Ratios become the subscripts: CH 3 Molecular Formula
Given that succinic acid is composed of 40.68% carbon, 5.08% hydrogen and 54.24% oxygen, find the empirical formula. ¨ ¨ Molecular Formula = (Empirical Formula)n n = molecular formula mass empirical formula mass
Example of Molecular Formula
¨ A sample contains 0.59g H and 9.40g O. The molecular mass of the cmpd is 34.0 g/mol. Find the empirical formula & the molecular formula for the molecule.