Formula Math & The Mole

I. Percent Composition –Gives the percent, by mass, of the elements in a compound –Grams of element x 100 grams of compound grams of compound

Two ways to calculate: 1) Chemical formula is given 2) Masses, but no formula, are given

Calculating percent composition when the formula is given: 1. Write the chemical formula. 2. Count the number of atoms of each element in the compound. 3. Multiply the # of atoms by the mass number from the periodic table (rounded to the nearest tenth) for each element. 4. Add answers together and round to the nearest tenth to determine grams of compound. 5. Use formula above to calculate % composition for each element. Round final answer to nearest hundredth.

Example # 1 What is the percent composition of water? One mole of water is grams. In that compound, there are two moles of H atoms and 2 x = grams. That's how many grams of hydrogen are present in one mole of water. There is also one mole of oxygen atoms weighing grams in the mole of water. To get the percentage of hydrogen, divide the by and multiply by 100, giving 11.19%. For oxygen it is ÷ = 88.81%.

Example # 2 Step 1: Determine the total mass of each element in the molar mass 1 mol Ca x (40.08gCa/1 mol Ca) = g Ca 2 mol N x (14.01gN/1 mol N) = g N 6 mol O x (16.00gO/1 mol O) = g O 1 mol of Ca(NO3)2 = g (molar mass) Step 2: Calculate percent by multiplying the mass ratio by 100% %Ca = (40.08gCa/164.1 Ca(NO3)2) x 100 = 24.42% %N = (28.02gN/164.1 Ca(NO3)2) x 100 = 17.07% %O = (96.00gO/164.1 Ca(NO3)2) x 100 = 58.50%

Example # 3 CO2 = 12 + (16 X 2) = 44 molecular weight (mass). C = 12/44 = 27.3% O = 32/44 = 72.7% CO2 = 12 + (16 X 2) = 44 molecular weight (mass). C = 12/44 = 27.3% O = 32/44 = 72.7%

Calculating percent composition when masses, but no formula are given: 1) 1) Find the sum of the masses of the elements that make up the compound. 2) 2) Divide the mass of each element by the total mass. 3) 3) Multiply by 100 to determine % composition of each element ( rounded to the nearest hundredth).

Example #1 what is the percent composition of a compound formed when grams of calcium reacts with grams of chlorine? what is the percent composition of a compound formed when grams of calcium reacts with grams of chlorine? = 75 g 27.07/75 =.36 x 100 = 36% calcium 47.93/75 =.64 x 100 = 64% chlorine

Example #2 What is the percent composition of a compound when 12.5 grams of carbon reacts with 16.3 grams of oxygen? = 28.8 grams 12.5/28.8 =.43 x 100 = 43% carbon 16.3/28.8 =.57 x 100 = 57% oxygen

Example #3 Seventy-five grams of nitrogen gas react with oxygen gas to form one hundred fifty grams of a compound. What is the percent composition of each element? 150 – 75 = 75 grams of oxygen 75/150 =.5 x 100 = 50% oxygen 75/150 =.5 x 100 = 50% nitrogen

In Chemistry, we commonly measure: –Mass (grams) –Volume (L or mL) –Particles (counted)

We can relate each of these measurements to a single quantity called the “Mole”.

The Mole is the SI unit for the “amount of something”

Why use it? To estimate the number of particles that are too small or too numerous to actually count.

Volume 1.0 mole = 22.4 Liters of a gas, at standard temperature and pressure (STP) Std. temp. = 0º C or 273K Std. pressure = 1 atmosphere (atm) or 760 mm Hg (Torr)

Particles 1 mole = x particles of a pure substance. * Element = atom * Molecular = molecule (mlc) *Ionic = formula unit (fmu)

Mass 1.0 mole = ____ grams of a pure substance. (“Molar Mass” of that substance)

Calculating molar mass: 1. Write the formula. 2. Count the number of atoms of each element in the formula. 3. For each element, multiply the # of atoms by its mass number (rounded to the nearest tenth) from the periodic table. 4. Add all products and round to the nearest tenth.

What is the Molar Mass of: Carbon tetrachloride 1. First write the correct molecular formula for carbon tetrachloride. CCl4 1. First write the correct molecular formula for carbon tetrachloride. CCl4 2. Add the molar masses of all the atoms in the molecule. Look at the periodic table and get these numbers. molar mass of C = g/mole molar mass of Cl = g/mole CCl4 has 1 mole of C and 4 moles of Cl. So the molar mass of CCl4 = 1(12.01 g/mole) + 4(35.45 g/mole) = g/mole

Oxygen Oxygen is a diatomic element: O x 2 = since you are looking for molar mass, units will be in grams per mole Oxygen is a diatomic element: O x 2 = since you are looking for molar mass, units will be in grams per mole

Aluminum carbonate Aluminum Carbonate is represented by the formula Al2(CO3)3. Two Al (2x 27g/mol)+three C (3 x 12g/mol)+ nine O (9 x 16) = 234g/mol!

III. Unit Conversions Remember: The mole is related to volume, particles, and mass. This allows you to convert from one unit to another.

To convert between units of the same substance: 1. 1.Read the entire problem Identify what is given (what you have) and what is unknown (what you want) Convert what is given into moles Convert from moles to the unknown Solve. Round to the correct number of significant figures and include proper units.

Remember…. 1 mole = x particles = 22.4 liters of gas, at STP ______ grams (Molar Mass)

Example On interwrite

E XAMPLE 2 How many moles of Magnesium is 3.01x10 22 atoms of magnesium? 3.01x10 22 atoms Mg X 1 mol Mg 6.02x10 23 atoms of Mg = 5.00x10 -2 mol Mg

Example: How many molecules are in grams of chlorine gas?

Example: What is the volume, in L, at STP, occupied by moles of sulfur dioxide gas?

Empirical Formula: Gives the lowest whole number ratio of elements in a compound From a molar perspective, the empirical formula represents the lowest whole number ratio of moles of each element in a compound Subscripts cannot be reduced

Molecular Formula: Is a whole number multiple of the empirical formula

Identifying Empirical vs. Molecular: Example # 1 : Ba 3 N 2 Example # 2 : Ca 3 (PO 4 ) 2 Example # 3: C 6 H 12 O 6 Example # 4 : H 2 O 2

Calculating Empirical Formula: 1.Change % sign to grams (assume a 100 gram sample). If grams are already given, go to step 2. 2.Convert from grams to moles. 3.Create a “mole ratio” by dividing all answers by the smallest answer. 1.If all answers are whole numbers, you are finished. Write formula for compound. 2.If all answers ARE NOT whole numbers, see step 4. 4.Multiply all answers from step 3 by “a number” that will give the lowest whole number ratio of elements. Write formula for compound.

Example # 1: What is the E.F. of a compound composed of 32.00% Carbon, 42.66% Oxygen, 18.67% Nitrogen, and 6.67% Hydrogen? Grams C = 100 * = 32.0 g; moles C = 32.0 g/(12.0 g/mole) = 2.67 moles Grams O = 100 * = g; moles O = g/(16.0 g/mole) = 2.67 moles Grams N = 100 * = g; moles N = g/(14.0 g/mole) = 1.33 moles Grams H = 100 * = 6.67 g; moles H = 6.67 g/(1.0 g/mole) = 6.67 moles Divide by the lowest number of moles (1.33): C: 2.67/1.33 = 2.0 O: 2.67/1.33 = 2.0 N: 1.33/1.33 = 1.0 H: 6.67/1.33 = 5.0 Empirical formula = C2H5NO2

Example # 2: What is the E.F. of a compound composed of 25.9% Nitrogen and 74.1% Oxygen? If 25.9% of the compound is nitrogen and 74.1% of the compound is oxygen, then a compound with a mass of 100 g has 25.9 g of nitrogen and 74.1 g of oxygen. This would mean that the ratio of nitrogen to oxygen is N 1.85 O We can divide each number in the ratio of N 1.85 O 4.63 by 1.85 to get N 1 O Since we cannot have 2.50 atoms of oxygen, we must multiply through each number by 2 to even it out, getting N 2 O 5 as our empirical formula.

Example # 3: What is the E.F. of a compound composed of 36.6 grams Carbon and 9.2 grams Hydrogen?

To determine the molecular formula, you need to know the empirical formula mass (you calculate) and the molecular formula mass (always given) M.F. mass = whole # multiple E.F. mass To determine the molecular formula, you need to know the empirical formula mass (you calculate) and the molecular formula mass (always given) M.F. mass = whole # multiple E.F. mass

Calculating molecular formulas: 1.Calculate E.F. mass 2.Divide M.F. mass by E.F. mass 3.Multiply E.F. by whole number answer from step 2. 4.Write M.F.

Example # 1: What is the M.F. of a compound with a mass of 92 grams and an E.F. of NO 2 ? Step 1: Determine the empirical formula molar mass N has 1 atom 1 X = g/mol O has 2 atoms 2X 16.00= 32.00g/mol g/mol Step 2: Divide the experimentally determined molar mass of succinic acid by the mass of the empirical formula to determine n n= Experimental Molar mass of NO 2 Molar mass of NO 2 n= 92g/mol g/mol n= Step 2: Multiple subscripts by 2 (n) N 2 O 4

Example # 2: What is the M.F. of a compound with the mass of 150 grams and an E.F. of CH 2 O? CH 2 O = (1 x g/mol) + (2 x g/mol) + (1 x g/mol ) = grams/ = 5 CH 2 O x 5 = C 5 H 10 O 5

Example # 3 What is the M.F. of a compound with a mass of 132 grams if that compound is composed of 54.6% Carbon, 13.6% Hydrogen, and 31.8% Nitrogen?

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