IIIIII Using The Mole

A. Percentage Composition n the percentage by mass of each element in a compound

 100 = A. Percentage Composition %Cu = g Cu g Cu 2 S  100 = %S = g S g Cu 2 S % Cu 20.15% S n Find the % composition of Cu 2 S. What does this calculation mean? If you have a sample of any amount of CU 2 S, about 80% (79.852%) of the mass is due to copper! Note that the total will add up to 100 % (80 % + 20%)

Example of Percent Comp. Cip..\..\tia demo.vob..\..\tia demo.vob

%Fe = 28 g 36 g  100 = 78% Fe %O = 8.0 g 36 g  100 = 22% O n Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. A. Percentage Composition

How many grams of copper are in a 38.0-gram sample of Cu 2 S? (38.0 g Cu 2 S)(0.7985) = 30.3 g Cu Cu 2 S is 79.85% Cu A. Percentage Composition

 100 = %H 2 O = g H g total 24.51% H 2 O n Find the mass percentage of water in calcium chloride dihydrate, CaCl 2 2H 2 O? A. Percentage Composition

B. Empirical Formula C2H6C2H6 CH 3 reduce subscripts n Smallest whole number ratio of atoms in a compound

B. Empirical Formula Using the mass (or % composition) of each element…. 1. Find the moles you have of each element. If % are given- they will add up to 100- so just assume you have a 100g sample- the percent will be the amount you have in grams. 2. Compare the relative amount of each element you have (find the ratio)... to do this, divide moles by the smallest amount to find subscripts/ratios this will be a 1: X ratio). 3. Realistic ratios must be whole numbers …so when necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

B. Empirical Formula n Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9g N 1 molN gN = 1.85 mol N present 74.1 g 1 mol g = 4.63 mol O present X X Find the moles of each

B. Empirical Formula n Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9g N 1 molN gN = 1.85 mol N present 74.1 g 1 mol g = 4.63 mol O present 1.85 mol = 1 N = 2.5 O X X Dividing the lower number by itself will give a 1 Dividing the higher number by the same number will give a 1:X ratio

B. Empirical Formula N 1 O 2.5 Is this a realistic ratio? Why or why not? We need to make the subscripts whole numbers  multiply by 2 N2O5N2O5

Bellwork: date: A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, &.675 g H. What is the empirical formula for this unknown compound? (A key to this problem will be handed out later)

C. Molecular Formula n “True Formula” - the actual number of atoms of each element in a compound CH 3 C2H6C2H6 empirical formula molecular Formula (one possible answer) ?

C. Molecular Formula 1. You will need the empirical formula. 2. Cal. the empirical formula mass (the GFM of the empirical formula). 3. You will need the molecular mass. (GFM of the real formula) 4.Divide the molecular mass (GFM) by the empirical mass (GFM). 5. Multiply each subscript by the answer from step 4 to get the molecular formula.

C. Molecular Formula n The empirical formula for ethylene is CH 2. Find the molecular formula if the molecular mass/(the GFM of the substance) is 28.1 g/mol? 28.1 g/mol g/mol = 2.00 empirical mass/GFM = 14.03g/mol CH 2 (how did we get this?) (CH 2 ) 2  GFM of simplest- ratio formula GFM of the actual formula of the compound C2H4C2H4