Chemistry Lesson: Calculating Molar Mass, Percent Composition, and Empirical Formulas, Slides of Chemistry

Download Chemistry Lesson: Calculating Molar Mass, Percent Composition, and Empirical Formulas and more Slides Chemistry in PDF only on Docsity! • Announcements • Molar Mass • Percent Composition • Empirical Formula • Chemical Formula Welcome to day 5! Announcements • Seminar Today – 4:00 this room • Information not covered in lecture are available elsewhere – Textbook – CAPA sets – CAPA tutors – Discussion sections – Office hours Percent Composition • Two types of percentage composition are logical – Mole percent and Mass percent • When text says “percent composition” they mean MASS – For example, strange fruit salad ingredients • 10 watermelons and 10 strawberries – By moles, salad is 50 % watermelon – By mass salad is 99 % watermelon • Let’s stick with our heme example C32H26FeN4O4 Composition • Mole percent – We have 32+26+1+4+4 = 67 total moles of components – Mole % Carbon (keep as many sig figs as you like) – Mole % Iron • Mass percent – Mass % Carbon (5 sig figs) – Mass % Iron (5 sig figs) 8.474776.0 67 32 ⇒= 5.10149.0 67 1 ⇒= 711.6565711.0 43.586 35.385 ⇒= 3524.9093524.0 43.586 845.54 ⇒= Recall that the chemical formula for heme is C32H26FeN4O4 O1mol g999.15 Hememol1 O mol4 N1mol g007.14 Heme mol1 N mol4 Fe1mol g845.55 Hememol1 Fe mol1 Hmol1 g0079.1 Heme mol 1 H mol26 C mol1 g011.12 Heme mol 1 C mol32 ×+×+×+×+× 54321 Heme1mol g996.63 Heme1mol g028.56 Heme1mol g845.55 Hememol1 g2054.26 Heme mol1 g352.384 ++++= Hememol1 586.43g = What is the mass percent of nitrogen in heme? 0% 0% 0% 0% 1. 10.913 % 2. 9.3524 % 3. 65.711 % 4. 9.5541 % Chemical Formula – glucose • With additional information we can find true chemical formula • For instance, maybe mass spectrum shows – Total mass of glucose roughly 180 g/mol • We know C:H:O must be 1:2:1 mol g mol g mol g mol g xxx 180999.1510079.12011.121 =×+×+× mol g mol gx 1800258.30 = 9948.5=x • Empirical formula is CH2O • Chemical formula is six times larger: C6H12O6 Empirical Formula of Magnesium Oxide • Start with known mass of Mg in a crucible • Add Oxygen (by burning) • Determine mass of MgxOy • Subtract to get mass of O • Example: – Mass of crucible 9.58 g – Mass of crucible + Mg 10.47 g – Mass of crucible + Mg + O 11.02 g Empirical Formula2 – Mag Oxide • Determine moles of Mg – Mass of Mg = 10.47g – 9.58g = 0.89g – Moles of Mg = 0.89g (1mol/24.305g) = 0.037 moles Mg • Determine moles of O – Mass of O = 11.02g – 10.47g = 0.55g – Moles of O = 0.55g (1mol/15.999g) = 0.034 moles O • Molar ratio? – 0.37:0.34 = 1.1 ≈ 1 • Therefore empirical formula is MgO