UNIT 8: THE MOLE (Counting Atoms)
Relating Mass to Number of Atoms How can the mass of an element from the periodic table be related to the actual number of atoms that are present in a sample? Use: The mole Avogadro’s number Molar mass
The Mole Definition-the amount of substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. Huh? If you weighed exactly 12 g of carbon-12 and counted every single atom in your pile, the number of atoms present is defined as 1 mol. Think of it like exactly 12 eggs are in 1 dozen.
Avogadro’s Number So how many carbon-12 atoms would be in that pile? 6.0221367 X 1023 Rounded to 6.022 X 1023 This number is “nicknamed” Avogadro’s number after the Italian scientist Amedeo Avogadro. A mole of any substance contains 6.022 X1023 particles, just like 1 dozen of any substance contains 12 particles.
If every person on earth (5 billion) counted continuously at 1 atom per second, it would take 4 million years for all atoms in one mole to be counted!
The mass of 1 mole of a pure substance is called the molar mass Equality that relates number of grams in a mole Molar mass is equal to the atomic mass of an element The unit for atomic mass is grams per mole (g/mol)
CONCEPTUAL WARM-UP (write your answer on a piece of scratch paper): The pictures below show a mole of water and a mole of gold. 1. Determine the mass (g) of each. Also determine the number of atoms in the gold and the number of molecules in water.
Now try this one… 1. How many grams AND how many atoms are in the mole of silver (left pic)? 2. How many grams and atoms are in the mole of carbon (right pic)?
Atoms, Moles, & Molar Mass can be used as conversion factors! When performing conversion between atoms, moles, and grams (molar mass), remember: A M G
When using masses from the periodic table, round to two decimals for all calculations!
Conversion Example 1 : M-G What is the mass, in grams, of 2.67 mol of the element copper, Cu? Given: 2.67 mol Cu Unknown: ? g Cu
Conversion Example 2: A-M How many moles of oxygen, O, are in 2,000,000 atoms of oxygen? Given: 2,000,000 atoms O Unknown: ? mol O
Conversion Example 3 A-M Convert 5.00 X 1023 molecules of chlorine gas to moles
Conversion Example 4 G-A How many atoms are in 5.8 gram of copper?
Conversion Example 5 A-G Determine the mass in grams of 2.058 X 1020 molecules of CO2
Molar Mass Used to Determine Percent Composition Molar mass-the mass of one mole of a compound 16
Why % composition? When a new compound is made or discovered, it is analyzed to determine its % composition. From this info, its empirical formula can be determined. 17
Percentage Composition Percentage by mass of each element in a compound Percentages should always total very close to 100! Always check! 18
Percentage Composition Example 1 Find the percentage composition of sodium nitrate. Formula of sodium nitrate: NaNO3 Molar mass of NaNO3=85.00 g/mol Total 100% 19
Percentage Composition Example 2 Find the percentage composition of silver sulfate. Formula of silver sulfate: Ag2SO4 Molar mass of Ag2SO4=311.81 g/mol Total 100.01% 20
Empirical Formula Simplified formula for a compound For an ionic compound, empirical formula is the compound’s formula unit because we always simplify the formula Calcium sulfide: Ca2S2CaS For a molecular compound, may not represent actual number of atoms present in a molecule Molecular formula: B2H6 Empirical formula: BH3 21
C2H6 CH3 Empirical Formula Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts CH3 22
How do you calculate empirical formula? 1. Write down the mass or % of each element given in the problem 2. Divide that number by the mass on the periodic table 3. Divide each number by the smallest number determined in step 2 4. IF you have a number in step 3 that isn’t a whole number (ex. 2.5), then multiply subscripts by 2, 3, or 4 to get whole #’s. 23
Empirical Formula Example 1 A compound consist of 75% carbon and 25% hydrogen by mass. Determine the empirical formula for this compound 75 g 1 mol 12.01 g = 6.24 mol N = 1 C 6.24 mol 25 g 1 mol 1.01 g = 24.75 mol O = 4 H 24
Empirical Formula Example 2 Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 1.85 mol 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O 25
Empirical Formula Example N1O2.5 Need to make the subscripts whole numbers multiply by 2 N2O5 26
CH3 C2H6 Molecular Formula “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? C2H6 molecular formula 27
How do you calculate molecular formula? 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3. 28
Molecular Formula Example The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol. empirical mass (CH2) = 14.03 g/mol 28.1 g/mol 14.03 g/mol = 2.00 (CH2)2 C2H4 29
General pathway to solve problems Mass of each element Percent composition of compound Empirical formula for compound Molecular formula for compound 30
Molecular Formula Example A sample of a compound with a molar mass of 34.00 g/mol is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula. 5.98 % H, 94.02 % O Empirical formula: HO Molecular formula: (HO)2=H2O2 31
Review of Topics Learned: 1. Calculate the molar mass of: A. magnesium sulfate B. silver (I) bromide 2. What is the percent composition of oxygen in calcium hydroxide? 3. What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen? 4. Given the following data, find the correct molecular formula 24.3 % C, 4.1 % H, 71.6 % Cl, molecular formula mass = 99.00 g.