How Are We Going to Determine the Empirical Formula of a Compound?
Chemistry 2 Jorge Nuñez
Ms. Principe / Period 2 February 8th, 2011
Aim: How are we going to determine the empirical formula of a compound?
Do Now:
1) What is the percent by mass of oxygen in H2C2O4?
H2 = 2(1) = 2 ; C2 = 2(12) = 24 ; O4 = 4(16) = 96 Gram Formula Mass = 90
Percentage of Oxygen = 6490 x 100 = 71%
2) Calculate the percentage of water:
-Mass of empty crucible and cover = 11.7g
-Mass of crucible, cover, and hydrate before heating = 14.9g
-Mass of crucible, cover, and anhydrous after heating = 14.53g
14.9g-11.7g = 3.2g (mass of hydrate) ; 14.9g-14.53g = .37g (mass of anhydrous)
Percentage of water: .373.2 x 100= 11.56%
HW: #1-3: On loose leaf, determine the empirical formula
1) 32.4% sodium; 22.5% sulfur; 45.1% oxygen
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2) 25.3% copper; 12.9% sulfur; 25.7% oxygen; 36.1% water I. Chemical Formula a. Empirical formula i. the simplest ratio of the atoms present in a molecule or compound ii. lowest number of atoms in a formula * Examples of empirical formula N2H4 = NH2; C6H6= CH 1. Steps: Based on % composition in each element, find: i. Mass ii. Mole iii. Divide each mole by lowest mole value iv. Whole number ratio of each element 2. Examples: v. 75% C, 25% H
Basis: 100g a. Mass: C = 75g, H = 25g b. Recall from Table T: mole = massatomic mass
Mole C = 75g12g/mole = 6.25 moles
Mole H = 251 g/mole = 25 moles c. Mole C = 6.25 moles6.25 moles = 1
Mole H = 25 moles6.25 moles = 4 d. CH4 vi. 13% Mg, 87% Br
Basis: 100g a. Mass: Mg = 13g, Br = 87g b. Recall from Table T: mole = massatomic mass
Mole Mg = 13g24g/mole = .54 moles
Mole Br = 87g80 g/mole = 1.1 moles c. Mole Mg = .54 mole.54 mole = 1
Mole Br = 1.1 mole.54 mole = 2 d.
Ms. Principe / Period 2 February 8th, 2011
Aim: How are we going to determine the empirical formula of a compound?
Do Now:
1) What is the percent by mass of oxygen in H2C2O4?
H2 = 2(1) = 2 ; C2 = 2(12) = 24 ; O4 = 4(16) = 96 Gram Formula Mass = 90
Percentage of Oxygen = 6490 x 100 = 71%
2) Calculate the percentage of water:
-Mass of empty crucible and cover = 11.7g
-Mass of crucible, cover, and hydrate before heating = 14.9g
-Mass of crucible, cover, and anhydrous after heating = 14.53g
14.9g-11.7g = 3.2g (mass of hydrate) ; 14.9g-14.53g = .37g (mass of anhydrous)
Percentage of water: .373.2 x 100= 11.56%
HW: #1-3: On loose leaf, determine the empirical formula
1) 32.4% sodium; 22.5% sulfur; 45.1% oxygen
-------------------------------------------------
2) 25.3% copper; 12.9% sulfur; 25.7% oxygen; 36.1% water I. Chemical Formula a. Empirical formula i. the simplest ratio of the atoms present in a molecule or compound ii. lowest number of atoms in a formula * Examples of empirical formula N2H4 = NH2; C6H6= CH 1. Steps: Based on % composition in each element, find: i. Mass ii. Mole iii. Divide each mole by lowest mole value iv. Whole number ratio of each element 2. Examples: v. 75% C, 25% H
Basis: 100g a. Mass: C = 75g, H = 25g b. Recall from Table T: mole = massatomic mass
Mole C = 75g12g/mole = 6.25 moles
Mole H = 251 g/mole = 25 moles c. Mole C = 6.25 moles6.25 moles = 1
Mole H = 25 moles6.25 moles = 4 d. CH4 vi. 13% Mg, 87% Br
Basis: 100g a. Mass: Mg = 13g, Br = 87g b. Recall from Table T: mole = massatomic mass
Mole Mg = 13g24g/mole = .54 moles
Mole Br = 87g80 g/mole = 1.1 moles c. Mole Mg = .54 mole.54 mole = 1
Mole Br = 1.1 mole.54 mole = 2 d.