Moles and Formula Mass

The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023 (on your reference sheet) There are exactly 12 grams of carbon-12 in one mole of carbon-12.

I didn’t discover it. It’s just named after me! Avogadro’s Number 6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855). I didn’t discover it. It’s just named after me! Note: If I sat for a portrait and this is how it came out, I’d be devastated. Amadeo Avogadro

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.941 g Li = g Li 45.1 1 mol Li

Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li = mol Li 2.62 6.941 g Li

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol 6.02 x 1023 atoms = atoms 2.07 x 1024 1 mol

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 1 mol Li 6.022 x 1023 atoms Li 6.941 g Li 1 mol Li (18.2)(6.022 x 1023)/6.941 = atoms Li 1.58 x 1024

Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 Empirical: H2O CH2O C12H22O11

Empirical Formula Determination Base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. Divide each value of moles by the smallest of the values. Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? (part 1) Divide each % by the element’s mass

Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 Empirical formula: C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

Combustion Analysis Device used to determine the mass percent of each element in a compound.

Can find % comp & emp form of an organic cpd of CHO by doing combustion analysis General equation for combustion…? All of the H bonds to O to form water, which goes into the water absorber All of the C bonds to O to form CO2, which deposits in the CO2 absorber Then find increase in mass of H2O and CO2 absorbers, that’s how much H2O and CO2 is gained O2 is the only added element b/c excess O2 is present in combustion

Ex) Combustion Analysis A 2.04g sample containing C, H, & O underwent combustion analysis. 4.48g of CO2 and 2.4g of H2O were produced. Find the empirical formula.

Process Use stoichiometry to find the grams of carbon and hydrogen produced. Subtract the total mass from the sum of the masses of carbon and hydrogen to find the mass of oxygen. Find moles of C, H, & O.

Find masses of C, H, & O Mass of C Mass of H Mass of O

Find moles of C, H, & O

Divide # moles of each element by smallest # calculated

Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255g of isopropyl alcohol produces 0.561g CO2 and 0.306g H2O. Determine the empirical formula of isopropyl alcohol.

When 0.1156g of a compound that contains only the elements carbon, oxygen, and hydrogen undergoes combustion analysis, it is found to produce 0.1638g of CO2 and 0.1676g of H2O. Determine the empirical formula of this compound.