FORMULA NUMBER OF MOLES GIVEN WEIGHT MOLAR MASS = GW MM n =

How many moles of carbon atoms are contained in 4 g of carbon? The atomic mass of carbon is 12 amu, therefore the molar mass of carbon is 12 grams. 4 g X 1 mole = 0.333 mole C 12 g CONVERSION FACTOR (from periodic table)

2. How many moles of water are there in 36 g of water? SOLUTION: The molecular mass of H2O is 18 amu. The molar mass of H2O is 18 g. 36 g X 1 mole = 2 moles H2O 18 g

3. Calculate the mass of aluminum carbonate (Al2(CO3)3) in 5 3. Calculate the mass of aluminum carbonate (Al2(CO3)3) in 5.85 moles of the compound. In Al2(CO3)3 , there are 2 Al, 3 C and 9 O Formula mass = (2 X 27 g/Al) + (3 X 12 g/C) + (9 X 16 g/O) = 234 g Molar mass = 234 g Mass of Al2(CO3)3 = 5.85 moles X 234 g = 1369 g

PERCENT COMPOSITION Percent by mass of each element present in the compound. % by mass of each element Total mass of element X 100% Total mass of compound =

What is the percentage composition of H2O? % H in H2O 2 X 1 g x 100% = 18 g = 11.11 % % O in H2O 1 X 16 g x 100% = 18 g 88.89% = TO CHECK = 100.00%

TYPES OF FORMULA

EMPIRICAL FORMULA - lowest whole number ratio of the elements in a compound STEPS 1. Given the % composition, change the percent to grams. 2. Change the grams to moles using the formula n = GW MM 3. Calculate the lowest whole number ratio of the moles.

Determine the empirical formula of a compound containing 74 Determine the empirical formula of a compound containing 74.19% Na and 25.81% O. Step 1. Change % to grams mass of Na = 74.19 g mass of O = 25.81 g Step 2. Solve for number of moles mole Na = 74.19 g = 3.2256 moles 23 g/mole Na mole O = 25.81 g = 1.613 moles 16 g/ mole O

EMPIRICAL FORMULA : Na2O Step 3. Divide all number of moles by the lowest number. number of atoms of Na 3.2256 moles = number of atoms of O 1.613 moles = 2 atoms 1.613 moles 1 atom 1.613 moles EMPIRICAL FORMULA : Na2O

Find the empirical formula of the compound that contains 33.32% Na, 20.30% N and 46.38% O. Step 1 – Change % to mass Na = 33.32 g O = 46.38 g N= 20.30 g Step 2 – Solve for number of moles mole Na = 33.32 g = 1.45 moles 23 g/Na mole N = 20.30 g = 1.45 moles 14 g/N mole O = 46.38 g = 2.90 moles 16 g/O

EMPIRICAL FORMULA : NaNO2 Step 3 – Divide all mole values by the lowest number of mole to get number of atoms. atom Na = 1.45 moles = 1 atom 1.45 moles atom N = 1.45 moles = 1 atom atom O = 2.90 moles = 2 atoms EMPIRICAL FORMULA : NaNO2

MOLECULAR FORMULA Indicates the actual number of atoms of each element present in the molecule. STEPS Find the empirical formula Find the empirical formula mass (EFM) Divide the given molecular mass (MM) by the EFM to get the multiplier of the empirical formula Multiply the subscript of each atom of the empirical formula by the multiplier to get the molecular formula.

Step 1 – Find the empirical formula N = 87.5 g H =12.5 g Find the molecular formula of a compound with a molar mass of 32 g/mole and percent composition of 87.5% N and 12.5% H. Step 1 – Find the empirical formula N = 87.5 g H =12.5 g Solving for number of moles: mole N = 87.5 g = 6.25 moles 14 g/N mole H = 12.5 g = 12.5 moles 1 g/H

EMPIRICAL FORMULA : NH2 Step 2. Get the EFM. Solving for the empirical formula atom of N = 6.25 moles = 1 atom 6.25 moles atom of H = 12.5 moles = 2 atoms EMPIRICAL FORMULA : NH2 Step 2. Get the EFM. EFM of NH2 = (1 N X 14 g) + (2 H X 1 g) = 16 g

Step 3. Divide the MM by the EFM to get the multiplier. multiplier = MM = 32 g = 2 EFM 16 g Step 4. Multiply the subscript of each atom in the empirical formula to get the molecular formula of the compound. N1X2 H2X2 = N2H4 MOLECULAR FORMULA