Empirical & Molecular Formulas Chapter 8

Empirical & Molecular Formulas Formaldehyde Acetic Acid Glucose colorless gas with a characteristic pungent odor known Human Carcinogen colorless liquid that when undiluted is also called glacial acetic acid main component of vinegar simple sugar product of photosynthesis the cell’s primary energy source CH 2 OCH 3 CO 2 H (CH 3 COOH) C 6 H 12 O 6 CH 2 O (CH 2 O) n (CH 2 O) 6 (CH 2 O) 2 (CH 2 O) n

Empirical Formula the formula of a compound that expresses the smallest whole-number ratio of the atoms present the simplest formula the formula that gives the composition of the molecules present Molecular Formula CH 2 OC 6 H 12 O 6

4 Magnesium oxide: Find the empirical formula Data: 0.519g Mg, 0.341g O 0.519g(1mol/24.31g) = mol Mg 0.341g(1mol/16.00g) = mol O Divide by the smallest mole value (in this case they are the same) mol Mg/ = 1mol Mg mol O/ = 1 mol O MgO

5 Using Percent Composition (by mass) to find empirical formula % Mg and O in MgO? 60.3% Mg and 39.7% O Treat the % values as mass values and change to moles 60.3g Mg and 39.7g O 60.3g(1mol/24.31g Mg) = 2.48mol Mg 39.7g(1mol/16.00g O) = 2.48mol O Divide by the smallest mole value 2.48mol Mg/2.48 = 1mol Mg 2.48mol O/2.48 = 1mol O MgO

6 An iron, oxygen, hydrogen compound was analyzed to be 52.2 % iron, 44.9 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. What is its name? Determine the % of hydrogen. ✓ 100% = 2.9% Treat the % values as mass values. ✓ Change to moles ‣ Fe: 52.2 g*1mole/55.8g = mole ‣ O: 44.9 g*1mole/16g = 2.81 mole ‣ H: 2.9 g*1mole/1.01g = 2.9 mole

7 An iron, oxygen, hydrogen compound was analyzed to be 52.2 % iron, 44.9 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. What is its name? Divide by the mole value by the smallest mole value ✓ Fe: 0.94 / 0.94 = 1 ✓ O: 2.81 / 0.94 = 3.0 ✓ H: 2.9 / 0.94 = 3.1 Round to the nearest whole number and Voilà: FeO 3 H 3 Consider the polyatomic ions can you find something in the chart?? ✓ Fe(OH) 3 Determine the name. ✓ iron (III) hydroxide

8 A compound is determined to be % calcium, % phosphorus and the rest oxygen. Determine the empirical formula. Then name this compound.

9 A compound is determined to be % calcium, % phosphorus and the rest oxygen. Determine the empirical formula. Then name this compound Determine the % of oxygen. ✓ 100% % % = 41.27% Treat the % values as mass values and change to moles ✓ Ca: 38.76*1mole/40g = mole ✓ P: 19.97*1mole/31g = 0.64 ✓ O: 41.27*1mole/16g = 2.58 mole Divide by the smallest mole value ✓ Ca: / 0.64 = 1.5 ✓ P: 0.64 / 0.64 = 1 ✓ O: 2.58 / 0.64 = 4 Multiply them all by a factor of 2 to get whole numbers and Voilà: Ca 3 P 2 O 8 Can we clean this up to look more like a familiar formula? Ca 3 (PO 4 ) 2 Calcium Phosphate

10 A carbon, hydrogen, oxygen compound was analyzed to be 40.0 % carbon, 53.0 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. The compound was further analyzed and has molar mass of 180 g/mole, determine the molecular formula. Determine the % of hydrogen. ✓ 100% - 53% - 40% = 7% Treat the % values as mass values. and change to moles ✓ C: 40g * 1mole/12g = 3.33 mole ✓ O: 53g * 1mole/16g = 3.31 ✓ H: 7g * 1mole/1.01g = 6.9 mole

11 A carbon, hydrogen, oxygen compound was analyzed to be 40 % carbon, 53 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. The compound was further analyzed and has molar mass of 180 g/mole, determine the molecular formula. Divide by the smallest mole value ✓ C: 3.33 / 3.31 = ✓ O: 3.31 / 3.31 = 1 ✓ H: 7 / 3.31 = 2 Rounding to the nearest whole number and Voilà - Empirical Formula: CH 2 O

12 A carbon, hydrogen, oxygen compound was analyzed to be 40 % carbon, 53 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. The compound was further analyzed and has molar mass of 180 g/mole, determine the molecular formula. Calculate the molar mass of the empirical formula ✓ CH 2 O (1) + 16 = 30 Divide the molar mass of the molecule by the molar mass of the empirical formula. ✓ 180 / 30 = 6 Drive the factor of 6 through the empirical formula to get the molecular formula ✓ C 6 H 12 O 6 is the molecular formula. This is actually the molecule. It’s sugar.

g phosphorus was burned in oxygen and the resulting compound weighed 7.8 g. Determine the empirical and molecular formulas, then name the compound. The molar mass is 284 g/mole. Calculate the mass of oxygen ✓ 7.8 g P ? O ? g P = 4.4 g O Determine the moles ✓ 4.4 g O * 1mol/16g = ✓ 3.4 g P * 1mol/31g = Divide by the smallest ✓ Oxy / = 2.5 ✓ P / = 1 multiply both by a factor (2x, 3x, maybe 4x) ✓ Oxy 2.5 x2 = 5 ✓ P 1 x2 = 2 Thus P 2 O 5

g phosphorus was burned in oxygen and the resulting compound weighed 7.8 g. Determine the empirical and molecular formulas, then name the compound. The molar mass is 284 g/mole. Empirical formula: P 2 O 5 Determine the molecular formula. Add the molar mass of the empirical formula ✓ 2(31) + 5(16) = 142 ✓ Divide into the molar mass given ✓ 284/142 = 2 ✓ Drive the 2 factor through the empirical formula ✓ P 4 O 10 name ???? ✓ tetraphosphorus decoxide

What is the empirical formula of a compound if a 50.0 g sample of it contains 9.12 g Na, 20.6 g Cr, and 22.2 g O? 15 Data: 9.11g Na, 20.6g Cr, and 22.2g O 9.11g(1mol/22.99g) = 0.396mol Na 20.6g(1mol/52.00g) = 0.396mol Cr 22.2g(1mol/16.00g) = 1.39mol O Divide by the smallest mole value 0.396mol Na/0.396 = 1mol Mg 0.396mol Cr/0.396 = 1 mol Cr 1.39mol O/0.396 = 3.5 mol O Multiply them all by a factor of 2 to get whole numbers and Voilà: Na 2 Cr 2 O 7 Name : sodium dichromate

Sulfadiazine, a drug used for the treatment of bacterial infections, analyzes to 48 % carbon, 4.0 % hydrogen, 22.4 % nitrogen, 12.8 % sulfur, and 12.8 % oxygen. The molecular mass is g/mole. Determine the empirical and molecular formulas 16 Percents to grams to moles 48g(1mol/12.01g) = 3.99mol C 4.0g(1mol/1.01g) = 3.96mol H 22.4g(1mol/14.01g) = 1.60mol N 12.8g(1mol/32.07g) = 0.399mol S 12.8g(1mol/16.00g) = 0.8mol O Divide by the smallest mole value 3.99mol C/0.399 = 10mol C 3.96mol H/0.399 = 9.9mol H 1.60mol N/0.399 = 4.01 mol N 0.399mol S/0.399 = 1 mol S 0.8mol O/0.399 = 2.0 mol O Empirical: C 10 H 10 N 4 SO 2 Molecular : same