EF, MF, Hydrates & % Comp notes

Empirical Formula E.F. is the smallest whole number mole ratios of the elements in a compound. In other words, half an atom does not exist and since elements are made of atoms when elements combine to form compounds, they combine in simple whole number ratios. 1 to 1, 1 to 2, 2 to 3, etc…

How to solve for an Empirical Formula Step 1: Find the number of moles for each element Step 2: Simplify the mole ratio by dividing ALL of the moles by the smallest number Step 3: (Only if needed) multiply to get a whole number mole ratio

Empirical Formula Example Problems Ex. 1) An ingredient in a hotdog consists of 5.801 g potassium, 2.078g nitrogen, and 7.121g oxygen. What’s in a hotdog?

Empirical Formula Example Problems Ex. 2) A car undergoes oxidation, the simplest definition of oxidation is the reaction with oxygen. It’s found to have 45.461g iron and 19.539g oxygen. What is this stuff?

Empirical Formula Example Problems Ex. 3) A cleaning bottle contains 82.24% nitrogen and 17.76% hydrogen. Which cleaning agent is this? *if the percentages add up to 100, assume you have 100g of the compound.

Molecular Formula M.F. is the actual formula of a molecular compound. Empirical formula contains the smallest possible whole numbers to describe the atomic ratio, while M.F. contains the correct number of atoms of each element in a molecule.

Molecular Formula M.F. C6H2O8N2 C6H6 C12H12 E.F.

Molecular Formula M.F. C6H2O8N2 C6H6 C12H12 E.F. C3HO4N

Molecular Formula M.F. C6H2O8N2 C6H6 C12H12 E.F. C3HO4N CH

Molecular Formula M.F. C6H2O8N2 C6H6 C12H12 E.F. C3HO4N CH CH

Molecular Formula M.F. C12H22O11 Ar4C4O8 H2O E.F.

Molecular Formula M.F. C12H22O11 Ar4C4O8 H2O E.F. C12H22O11

Molecular Formula M.F. C12H22O11 Ar4C4O8 H2O E.F. C12H22O11 ArCO2

Molecular Formula M.F. C12H22O11 Ar4C4O8 H2O E.F. C12H22O11 ArCO2 H2O Sometimes the molecular formula and empirical formula are the same…

Molecular Formula Once you know empirical formula, you can determine the molecular formula. Compare molar mass for each (by dividing) Molar mass of compound = n Molar mass of empirical formula (Copy everything in yellow)

Molecular Formula If masses are equal, they are the same formula If not equal, multiply each subscript in empirical formula by the factor (n) to get molecular formula Molecular Formula = n(Empirical Formula)

How to solve for a Molecular Formula Step 1: If not already given, you must 1st determine the empirical formula Step 2: Add up the molar mass for empirical formula Step 3: Compare the molar mass of the empirical formula to the molar mass of the actual compound

Molecular Formula Example Problem Ex) A compound with a mass of 42.08 amu is found to be 85.64% carbon and 14.36% hydrogen. Find its molecular formula.

Hydrates Sometimes ionic substances have water molecules in them. These substances are called hydrates. These hydrates will absorb specific ratios of water molecules. Formulas for hydrated ionic compounds are written with the water included.

Hydrates Hydrated copper(II) sulfate is CuSO4·5H2O. There are 5 moles of water for every 1 mole of cupric sulfate. Let’s find the molar mass for CuSO4·5H2O and name it.

Hydrate Example Problem Heating the hydrated sample can drive off the water and leave the salt behind Ex) After heating 67.5g of hydrated calcium sulfate, CaSO4·xH2O, you have 53.4g of pure CaSO4. What is the formula and name of the hydrate?

Percent Composition Percent Composition is a statement of the relative mass each element contributes to the mass of the compound as a whole. A chemist often compares the percent composition of an unknown compound with the percentages calculated from an assumed formula. If the percentages agree, it helps confirm the identity of the unknown compound.

Percent Composition Shows parts of the whole The percent by mass of each element in the compound – must add up to 100% Is constant for each compound – Law of Definite Proportions Can be used to identify unknown substances

Percent Composition Shows parts of the whole The percent by mass of each element in the compound – must add up to 100% Is constant for each compound – Law of Definite Proportions Can be used to identify unknown substances

Percent Composition Mass of element X 100 Mass of compound

Percent Composition Example Problems Ex 1) Find the percent composition of each element in sodium chloride, ______

Percent Composition Example Problems Ex 1) Find the percent composition of each element in sodium chloride, NaCl

Percent Composition Example Problems Na: mass of Na x 100 = 22.990 x 100 = 39.337 % Na mass of NaCl 58.443 Cl: mass of Cl x 100 = 35.453 x 100 = 60.663 % Cl mass of NaCl 58.443 * percentages must add up to 100%

Percent Composition Example Problems Ex 2) Find the percent composition of each element in aluminum sulfate, __________

Percent Composition Example Problems Ex 2) Find the percent composition of each element in aluminum sulfate, Al2(SO4)3