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Weds March 5 , 2014

Weds March 5 , 2014. Due: HW 7C, Lab Reports Today: Determining Chemical Formulas Empirical Formulas Molecular Formulas Friday: Magnesium Oxide Lab Bring your own goggles if you don’t want to wear the ones in the classroom set! . Necessary skills: . Multiplication Division

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Weds March 5 , 2014

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  1. Weds March 5, 2014 • Due: HW 7C, Lab Reports • Today: Determining Chemical Formulas • Empirical Formulas • Molecular Formulas • Friday: Magnesium Oxide Lab • Bring your own goggles if you don’t want to wear the ones in the classroom set!

  2. Necessary skills: • Multiplication • Division • Convert mass to moles

  3. Calculation of Chemical Formulae • A Molecular formula includes the symbol of elements and the number of atoms of each element in that molecule • Empirical formula includes symbols of elements in compounds with subscripts that show the smallest possible whole numbers that describe the atomic ratio

  4. Determining Chemical Formula • Ionic compounds – formula unit IS ALREADY the smallest whole-number ratio • Molecular compounds (covalent bonds) – molecule is not always the smallest whole number ratio!

  5. Methane • Molecular formula: CH4 • Empirical formula: • CH4

  6. Ethane • Molecular formula: C2H6 • Empirical formula: • CH3

  7. Water • Molecular formula: H2O • Empirical formula: • H2O

  8. Benzene • Molecular formula: C6H6 • Empirical formula: • CH

  9. Calculation of Empirical Formula • 1. Determine Mass Composition • If given percentages: Use percentage composition to convert to a mass composition (assume 100 g sample so percent is equal to the mass in g) • If given mass: Skip this step • 2. Convert mass to moles for each element • 3. Find the smallest whole-number ratio by dividing each number of moles by the smallest number • 4. These whole numbers are the subscripts in your compound

  10. Example 1 – Step 1 • A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. • 1. Use % composition to get to mass composition • 32.38% Na = 32.38 g Na • 22.65% S = 22.65 g S • 44.99% O = 44.99 g O

  11. Example 1 – Step 2 • A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. • 2. Convert mass to moles • 32.38 g Na / (22.989 g/mol) = 1.408 mol Na • 22.65 g S / (32.065 g/mol) = 0.706 mol S smallest • 44.99 g O / (15.999 g/mol) = 2.812 mol O

  12. Example 1 – Step 3 • A compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula. • 3. Find smallest ratio by diving by smallest number of moles • Na: 1.408 mol / 0.706 = 1.99 = 2 • S: 0.706 mol / 0.706 = 1 • O: 2.812 mol/ 0.706 = 3.98 = 4 • 4. Write empirical formula • Empirical formula = Na2SO4 These numbers become the subscripts in the empirical formula.

  13. Example 2 – Step 1 • A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. • 1. Use % composition to get to mass composition (you are already there!) • 4.43 g P • 5.72 g O

  14. Example 2 – Step 2 • A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. • 2. Convert mass to moles • 4.43 g P / (30.974 g/mol) = 0.143 mol P smallest • 5.72 g O / (15.999 g/mol) = 0.358 mol O

  15. Example 2 – Step 3 • A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. • 3. Find smallest ratio by diving by smallest number of moles • P: 0.143 mol / 0.143 = 1 x 2 = 2 • O: 0.358 mol O / 0.143 = 2.5 x 2 = 5 • THEY MUST BE WHOLE NUBMERS! YOU CAN’T HAVE HALF AN ATOM! Multiply by 2 (or 3, if the decimal is .333 or .666) to get whole numbers

  16. ROUNDING HELPS • If the number ends in .98, .99, .01, or .02  Round to nearest whole number • If the number ends in .50, .51, .52, .49, .48, etc  Round to .5 then multiply by 2 • If the number ends in .33, .32, .34, etc  Round to .3 then multiply by 3 • If the number ends in .66, .65, .64, etc  Round to .6 then multiply by 3

  17. Example 2 – Step 4 • A compound contains 4.43 g phosphorus and 5.72 g oxygen. Find the empirical formula. • 4. Write empirical formula • P = 2 • O = 5 • Empirical formula: P2O5 These numbers become the subscripts in the empirical formula.

  18. Calculation of Molecular Formula • If you know the empirical formula and the molecular mass of a compound, you can determine the molecular formula. • There are four pieces of information. If you know three, you can solve for the fourth. • 1. Empirical formula • 2. Molecular formula • 3. Empirical mass (is the mass of the empirical formula) • 4. Molecular mass (is the mass of the molecular formula)

  19. Calculation of Molecular Formula 1 16.05 16.05 15.04 2 30.08 18.02 1 18.02 13.01 78.12 6

  20. What is x? The relationship between masses (x) will tell you the relationship between formulas. Find x by: x = molecular mass empirical mass

  21. Steps • 1. Calculate the empirical mass (like you are used to doing.) • 2. Solve for x using molecular mass (given in the problem) and empirical mass. • 3. Multiply the empirical formula by x x(empirical formula) = molecular formula

  22. Example 1 • Determine the molecular formula of the compound with an empirical formula of CH and a molecular mass of 78.110 amu. • Empirical formula= CH • Molecular formula= ?? • Empirical mass= mass C + mass H = 12.01 + 1.01 = 13.02 amu • Molecular mass= 78.110 amu • x = molecular mass = 78.110 = 5.999 formula mass 13.02 • 6(CH) = C6H6

  23. Example 2 • A sample of a compound with a molar mass of 34.00 g/mol consists of 0.44 g H and 6.93 g O. Find its molecular formula. • Empirical formula: Must determine from data given in the problem (See p.6 of notes) • Molecular formula: ?? • Empirical mass: Can determine from empirical formula • Molecular mass: 34.00 • H: 0.44 g H x (I mol H) = 0.436/0.433 = 1 (1.01 g H)  HO (empirical formula) • O: 6.93 g O x (I molO) = 0.433/0.433 = 1 (16.00 g O) • x = molecular mass = 34.00 = 1.99 empirical mass HO (16.00 + 1.01) • 2(HO) =H2O2

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