Chapter 1: Stoichiometric relationships

1.2: The mole concept Note: You should review naming Use the chemistry help page on website for some review videos Click image!

Avogadro’s number and relative atomic mass Avogadro’s # (L or NA) = 6.02x1023 = 1 mol Relative Atomic mass (Ar) The weighted average of the different isotopes that exist in nature Has NO units! Ar= weighted average of 1 atom of element 1/12 mass of one atom of Carbon-12 Isotope Carbon-12 is the international reference point for measuring atomic (and therefore molecular and molar) masses Carbon-12 has a value of exactly 12 1/12 of an atom of Carbon-12 has a value of exactly 1

Mole conversion examples How many formula units are in 0.5 moles of NaCl? How many grams are in 5.0 moles of hydrogen perioxide? How many moles are in 0.3 g of Iron(IV)Oxide? How many atoms are in 2.4 moles of water?

Relative molecular/formula mass and molar mass Relative Molecular/formula mass (Mr) The sum of the weighted average atomic masses in a compound Has NO units! Mr of H2O = H + H + O where H = 1.01 and O = 15.99 Mr of H2O = 1.01+1.01+15.99 = 18.01 Molar mass (M) The mass of 1 mole of a substance Units are g mol-1 n = number of moles (mol) m = mass in grams (g) M= molar mass (g mol-1)

Mole conversion examples How many moles are in 52 grams of ice? How many grams are in 3 moles of dry ice? What is the molar mass of a sample that is has 5 grams and 0.25 moles? If the previous sample was made of only one atom, what might the identity of that element be?

Mole ratios – compounds 1 mole of a substance is 6.02x1023 So if we have 1 mol of Na2SO4, then we have: 1 mol of S atoms 2 mol of Na atoms 4 mol of O atoms If we have 2 mol of H2O, then we have: 4 mol of H atoms 2 mol of O atoms If we have 0.5 mol of FeO2, then we have: 0.5 mol of Fe atoms 1 mol of O atoms

Mole ratios – equations Use the following equation: 2H2 + O2  2H2O If we have 1 mol of O2: We need 2 mol of H2 to make 2 mol of H2O If we have 2 mol of O2: We need 4 mol of H2 to make 4 mol of H2O If we have 1 mol H2: We need 0.5 mol of O2 to make 1 mol of H2O If we need to make 5 mol H2O: We need 5 mol H2 and 2.5 mol O2

Empirical formulas The simplest ratio of atoms in a compound MgO CH4 C2H6O7 Empirical means “experimental” What is the empirical formula for: C2H6 Ag2N6O2 N3O9 Si4O10

Empirical formulas How to calculate Empirical Formulas from experimental data: A sample contains 63.5 g Ag, 8.2 g N and 28.3 g O. What is the empirical formula of the compound?

Empirical formulas – easy Tetraethyl lead, a gasoline additive, is found to contain 38.43 g lead, 17.83 g carbon, and 3.74 g hydrogen. What is its empirical formula?

Empirical formulas – medium An oxide of chromium is found to have the following % composition: 68.4% Cr and 31.6% O. What is the empirical formula?

Empirical formulas – Hard 200.00 g of an organic compound is known to contain 83.884 g of carbon, 10.486 g of hydrogen, 18.640 g of oxygen and the rest is nitrogen. What is the empirical formula of the compound?

Molecular formulas from empirical formulas – easy An empirical formula is determined to be NO2 and the molar mass is 92.5 g mol-1. What is the Molecular Formula?

Molecular formulas from empirical formulas – medium A sample with molecular mass of 170 g mol-1 contains 63.5 g Ag, 8.2 g N and 28.3 g O. What is the molecular formula of the compound?

Molecular formulas – hard A 10g sample of a compound contains 4.00g C, 0.667g H, and the rest is Oxygen. Find the empirical and molecular formulas. The MW is 180.

Empirical formulas: hydrates What happens when something hydrated is dehydrated? Hydrate: compound that has water bound Ex: copper sulfate pentahydrate (CuSO4∙5H2O) and copper sulfate trihydrate (CuSO4∙3H2O) Ex: Na2CO3∙10H2O MgSO4∙7H2O Anhydrous: water has been removed from hydrate Ex: anhydrous copper sulfate (CuSO4) Ex: Na2CO3 MgSO4

Empirical formulas: hydrates - easy barium chloride dihydrate = ___________ What is the percent composition of water in the compound above? 36.04 / 246.4 x 100 = 14.6% Magnesium sulfate heptahydrate = _____________ If 125 grams of the compound above is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain? 100 – 51.14 = 48.86 % magnesium sulfate 0.4886 x 125 = 61.1 grams

Empirical formulas: hydrates – medium A 20g sample of a hydrate of nickel sulfate (NiSO4) lost 9.63g of water when heated. Determine the hydrate’s formula.

Empirical formulas: hydrates - hard A hydrate containing copper, sulfur, oxygen, and water lost 9g upon heating. Originally the hydrate had weighed 25g. Analysis of the anhydrous substance revealed that the 6.4g of Cu, 3.2g of S, and 6.4g of O were present. Find the formula of the hydrate. Ermahgerd!!!!.....whaaaat???

A hydrate containing copper, sulfur, oxygen, and water lost 9g upon heating. Originally the hydrate had weighed 25g. Analysis of the anhydrous substance revealed that the 6.4g of Cu, 3.2g of S, and 6.4g of O were present. Find the formula of the hydrate.

A hydrate containing copper, sulfur, oxygen, and water lost 9g upon heating. Originally the hydrate had weighed 25g. Analysis of the anhydrous substance revealed that the 6.4g of Cu, 3.2g of S, and 6.4g of O were present. Find the formula of the hydrate.

Empirical formulas: Percentage composition Percentage composition by mass can be calculated from the empirical formula! What percentage by mass of hydrogen (H) is in water? What is the percentage by mass of N and O in NO3? What is the percentage by mass of H, S and O in H2SO4?

Empirical formulas: Combustion analysis What are the products of the combustion of a hydrocarbon? CH4 + O2  CO2 + H2O For the amount of carbon in the original hydrocarbon, this is equal to the number of CO2 molecules formed. For the amount of hydrogen in the original hydrocarbon, the water reflects this value.

A 15.25 gram sample of an organic compound was combusted in oxygen producing 34.71 grams of CO2 and 14.20 grams of H2O. In addition it was found that the compound contained 27.59% oxygen. What is the empirical formula of the compound?

A 12 g sample of a smelly compound was tested by combustion analysis A 12 g sample of a smelly compound was tested by combustion analysis. The products were 21.41 g of CO2, 14.59 g of H2O, and 17.51 grams of N2O5. Further analysis showed that oxygen was NOT present in the molecule. What is the empirical formula of the compound?

Empirical formulas: Combustion analysis A 0.225 g sample of a hydrocarbon that contains oxygen, caproic acid, is combusted producing 0.512g CO2 and 0.209 g H2O. The molar mass of the acid is 116g mol-1 what is the molecular formula for caproic acid?