1. Chapter Calculations with Chemical Formulas and Equations Chemistry 1061: Principles of Chemistry I Andy Aspaas, Instructor

2. Molecular weight and formula weight Molecular weight: sum of atomic weights for all atoms in a substance –MW of H 2 O = 2(1.008 amu) + (16.00 amu) = 18.016 amu Formula weight: same as above, but for any compound, molecule or not –FW of NaCl = (22.99 amu) + (35.45 amu) = 58.44 amu

3. Mass and moles Mole: quantity of a substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon-12 –That number = Avogadro’s number, 6.02 x 10 23 –Simply a quantity like pair, dozen, gross Molar mass: mass of 1 mole of a substance –Equal to formula weight of that substance! –One mole of oxygen has a mass of 16.00 g –One mole of H 2 O has a mass of 18.016 g

4. Mass and moles Mole calculations –Use atomic, molecular, or formula weight as a conversion factor in dimensional analysis to convert between mass and moles, and vice-versa –How many moles in 100.0 g Fe? 100g of H 2 O? –What is the mass of 4.72 moles of Mg? 3.25 moles of glucose (C 6 H 12 O 6 )?

5. Molecules and moles Use Avogadro’s number as a conversion factor to convert between number of particles and moles, and vice versa 6.02 x 10 23 particles = 1 mole Combine the two methods to convert mass to number of particles.

6. Determining mass percentage from a formula Mass percentage: mass of certain constituent divided by mass of whole, expressed as percentage Mass % from formula: –Assume you have 1 mol of the substance, what mass do you have? –Use formula to determine number of moles of each element present –Convert moles of element to mass –Divide mass of element by mass of compound and multiply by 100%

7. Elemental analysis Determines amount of C, H, and O in a compound that contains only those elements Compound is burned in an oven, amounts of CO 2 and H 2 O given off are recorded Since 1 mol of CO 2 contains 1 mol C atoms, the mass of carbon in the substance can be found 1 mol H 2 O contains 2 mol H atoms - mass of H can be found Mass % of C and H found O mass % is what’s left over

8. Determining empirical formulas from % composition Empirical formula: simplest formula of a substance that has all integer whole-number subscripts If given % compositions of elements, assume you have 100.0 g of the substance –14.0% of an element becomes 14.0 g of that element Convert grams to moles for each element Divide all mole numbers by the smallest one If not all answers are integers (within experimental error), multiply to make them all the smallest possible integers

9. Molecular formula from empirical formula Molecular formula is some multiple of the empirical (simplest) formula –Empirical formulas of acetylene (C 2 H 2 ) and benzene (C 6 H 6 ) are both CH Molecular weight = n x empirical formula weight n = molecular weight / empirical formula weight Multiply all subscripts in empirical formula by n

10. Stoiciometry Chemical equations can be interpreted as either individual molecules or numbers of moles N 2 (g) + 3H 2 (g) ---> 2NH 3 (g) –1 N 2 molecule reacts with 3 H 2 molecules to make 2 NH 3 molecules –Or… 1 mol N 2 reacts with 3 mol H 2 to make 2 mol NH 3 Use coefficients as conversion factors when converting moles of any constituent in a chemical reaction to moles of any other constituent

11. Masses in chemical equations How many grams of a product will be yielded from a given mass of reactant? Do same process as before, but convert masses to moles first, and moles back to masses –Mass NH 3 --> mol NH 3 --> mol H 2 --> mass H 2

12. Limiting reactant Often times, one reactant will be in excess of the other Number of moles of product is determined by starting moles of limiting reactant Determine number of moles of each reactant, and calculate how many moles of product each would make The reactant that makes the least amount of moles of product is the limiting reactant

13. Theoretical yield Theoretical yield: maximum amount of product that can be obtained by a reaction from given amounts of reactants In actuality, less product is often obtained –Losses due to side-reactions –Losses due to incomplete separations Percentage yield = (actual yield / theoretical yield) x 100%