Math in Chemistry

Let’s start off with something familiar Remember the Beanium Lab Let’s start off with something familiar Remember the Beanium Lab??? We figured out percent abundance… Took the number of one type of bean and put it over the total number of beans Well this is really quite similar

Percent Composition Used to figure out chemical formulas The percentage by mass of each element in a compound

Percent Composition Two types of Problems Masses are Given No masses are Given

Masses are Given Steps to Solve Problem: Add given masses to get total mass for the compound Divide individual mass by the total mass Multiply by 100 to get the percent

Example: Masses Given Find the percent composition of a sample that is 30g Mg and 8.0g O. %Mg = 30g/38g X 100 = 79% Mg %O = 8.0g/38g X 100 = 21% O

No Masses Given Steps to Solve Problem: Assume you have 1 mole of the compound Calculate the molar mass of each element in the compound by multiplying the subscript by the molar mass of the element Divide the molar mass for the element by the total molar mass of the compound Multiply by 100 to get the percent

Example: No Masses Given Calculate the percent composition of oxygen in water Formula: H2O H: (2)(1.00794)= 2.01588 g/mol O: (1)(15.9994) = 15.9994 g/mol Total molar mass (2.01588 + 15.9994)= 18.0153 g/mol %O = 15.9994/18.0153 X 100 = 88.8101%

Your Turn: Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. C: (3)(12.0107)=36.0321g/mol H: (8)(1.00794)=8.06352g/mol (36.0321+8.06352)=44.0956g/mol %C = 36.0321/44.0956 X 100 = 81.7136% %H = 8.06352/44.0956 X 100 = 18.2865%

Smallest whole number ratio of the atoms of the elements in a compound Empirical Formula Smallest whole number ratio of the atoms of the elements in a compound C2H6 reduce subscripts CH3

Empirical Formula Steps to solve problem Find mass (or %) of each element. Find moles of each element (divide given mass by molar mass) Divide answers by the smallest # to find subscripts When necessary multiply subscripts by 2,3, or 4 to get whole #’s.

Example: A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Step 1: percent to mass (assume 100.0g) N: 25.9% = 25.9g O:74.1%=74.1g Step 2: mass to mole 25.9g N 1mol = 1.85 mol N 14.0067g N 74.1g O 1mol = 4.63 mol O 15.9994g O Step 3: divide by smallest N: 1.85/1.85 = 1 O: 4.63/1.85 = 2.5 Step 4: multiply until whole Since O came out to 2.5 we have to multiply both by 2 to get rid of the decimals N: 1 x 2 = 2 mol N O: 2.5 x 2 = 5 mol O The Empirical Formula is N2O5

Your turn: 1,6-diaminohexane is used to make nylon. What is the empirical formula of this compound if it is 62% C, 13.8% H, and 24.1% N? Answer: C3H8N

Molecular Formula The “true formula” or actual number of atoms in a compound Can be either the same as its experimentally determined empirical formula, or it is a simple whole number multiple of its empirical formula

Molecular Formula Steps to solving problems Find the empirical formula Find the empirical formula mass Divide the molecular mass by the empirical mass Multiply each subscript by the answer from step 3

Example: Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Step 1: Find empirical formula Empirical formula given: CH4N. Step 2: find empirical formula mass C: (1)(12.0107) = 12.0107 H: (4)(1.00794) = 4.03176 N: (1)(14.0057) = 14.0057 12.0107+4.03176+14.0057 = 30.0482g/mol Step 3: divide molecular mass by empirical mass 60.0g/30.0482 = 2.00 Step 4: multiply each subscript by the answer from #3 C2H8N2

Your Turn: The compound methyl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and 31.4% O and its molar mass is 102 g/mol. What is its empirical formula? What is its molecular formula?

One more: You find that 7.36g of a compound has decomposed to give 6.93g of oxygen. The only other element in the compound is hydrogen. If the molar mass of the compound is 34.0 g/mol, what is its molecular formula?