formula weight (molecular weight) - sum of the atomic weights of all atoms in the species

Determine the MW of SCl2

Determine the MW of calcium nitrate

mole – a quantity that contains 6.022 x 1023 particles

Avogadro’s #, N – 1 mole of anything = 6.022 x 1023 particles

Amedeo Avogadro 1776 - 1856

Lorenzo Romano Amedeo Carlo Bernadette Avogadro di Quaregna e Cerreto

How many grams does 1 mole of lead, Pb weigh ?

molar mass – one mole (6.022 x 1023) of any substance is equal to the atomic weight of that substance (units are g/mole)

How many grams of Al are in 1 mole of Al atoms ? 26.981538 g Al

What is the mass of a single Al atom ?

26.981538 amu

What is the molar mass of Al ?

26.981538 g/mole

I have 26.981538 g Al. How many atoms of Al do I have ? 6.022 x 1023 atoms Al

Which container has more atoms:207.2 g Pb or 26.981538 g Al ? same

How many atoms are in 207.2 g Pb ?

6.022 x 1023 atoms Pb

Calculate the # moles in 4.558 g K

Calculate the # moles in 38.620 g H2O

Calculate the # grams in 3.97 moles CO2

Calculate the # molecules in 3.704 g H2O

Calculate the # of moles of carbon, C in 5.05 moles CO2

Calculate the # moles Cl- anions in6.314 moles CaCl2

Calculate the # of hydrogen atoms in 14.5 g CH3OH

Percent composition = mass of element total mass species

x 100

Calculate the percentage composition of each element in chromium(III) oxide

empirical formula – formula containing the smallest, whole # ratio of atoms in a molecule

Empirical Formula’s in Calculations

1. Calculate the moles of each substance

2. Divide each moles by the smallest number to get the empirical formula

A compound was analyzed and found to contain 7.889 g sulfur and 11.808 g oxygen. What is the empirical formula of the compound ?

An oxide of gold, Au, is analyzed and found to be 89.11% Au (by wt.) and 10.89% oxygen (by wt.). What is the empirical formula of compound ?

* assume 100 g sample

How and When to round during empirical formula calculations

D = 1Q ≈ 1.5

D = (1)2 = 2Q ≈ (1.5)2 ≈ 3 D2Q3

D = 1Q ≈ 1.33

D = 1Q = 1.94748

D = (1)3 = 3Q ≈ (1.33)3 ≈ 4

D3Q4

≈ 2 DQ2

Empirical formula vs. Molecular formula

1. Calculations always reveal empirical formula

2. Must be given the molecular weight, MW, to determine actual molecular formula

3. Ratio the actual MW vs. the MW of empirical formula

C 2 H 3 O 1

emp formC2H3O

Is C2H3O the molecular formula ?

Given: the molecular weight, MW for the actual compound is 129.1 g/mole

MW43.045 g/mole

129.1 g/mole 43.045 g/mole

≈ 3 3(C2H3O)

C6H9O3molecular formula

hydrates – salts with loosely held H2O’s of hydration

Na2SO4·7H2O sodium sulfate heptahydrate

Mg(NO2)2 ·3H2O magnesium nitrite trihydrate

CaCl2 exists as a hydrate. 6.267 g of hydrated calcium chloride was heated leaving 4.731 g of dehydrated CaCl2. Determine:

1.the %water (by wt.) in the hydrated salt

2.the empirical formula of the hydrated salt

balanced chemical reaction – same number of atoms of each element on either side of the arrow

reactants – are on left side of the arrow; stuff being consumed in reaction

products – are on right side of the arrow; stuff being formed in reaction

arrow, – means equals, or produces, or gives

combustion rxn – when something burns and consumes oxygen, O2

Balance reactions by adding coefficients to get the same number of each atom of each element on either side of the arrow

Balancing Hints

1. Inspect the unbalanced rxn for 10 sec

2. Leave the free, uncombined elements alone until the end

3. Work with pencil so you can erase mistakes

4. Practice lots of hw problems

stoichiometry – a quantitative relationship between reactants and products via moles

How many moles of H2O are obtained from the combustion of 3.5 moles C3H8O ?

How many moles of O2 are required to completely combust 12.1 moles C3H8O ?

How many g of CO2 can be obtained by combusting 3.70 moles of C3H8O ?

Isopropyl alcohol, C3H8O was burned and 12.5 g O2 was consumed. How many g H2O was produced ?

Limiting reagent – reactant that runs out first

Excess reagent – reactant left over after the rxn is finished

Theoretical yield – maximum obtainable product (based on the limiting reagent)

Percent yield = amount of product actually obtainedtheoretical yield of product

x 100

15.0 g C3H8 was combusted with 15.0 g O2.How many g of H2O are theoretically possible ?

1. Write the rxn and balance it

2. Calculate the theoretical yield starting with each reactant. The smallest value obtained is the theoretical yield and is thus the one based on the limiting reagent

15.0 g C3H8 was combusted with 15.0 g O2 and 5.84 g H2O is obtained. What is the %yield of H2O ?