Stoichiometry Review Spring Final Exam

PbO 2 HNO 3 Pb: 1 (207.2 g) = g O: 2 (16.0 g) = 32.0 g g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0 g 63.0 g O: 3 (16.0 g) = 48.0 g Molar Mass the mass of one mole of a substance

percentage composition: the mass % of each element in a compound Find % composition. PbO 2 (NH 4 ) 3 PO 4 % of element = g element molar mass of compound x g Pb g : = 86.6% Pb 32.0 g O g = 13.4% O 31.2 g P g = 20.8% P 64.0 g O g = 43.0% O 42.0 g N g = 28.2% N 12.0 g H g = 8.1% H : : : : : (see calcs above)

Finding an Empirical Formula from Experimental Data 1. Find # of g of each element. 2. Convert each g to mol. 3. Divide each “# of mol” by the smallest “# of mol.” 4. Use ratio to find formula. A compound is 45.5% yttrium and 54.5% chlorine. Find its empirical formula. YCl 3

(How many empiricals “fit into” the molecular?) To find molecular formula… A. Find empirical formula. B. Find molar mass of empirical formula. C. Find x = MM molecular MM empirical D. Multiply all parts of empirical formula by x.

A carbon/hydrogen compound is 7.7% H and has a molar mass of 78 g. Find its molecular formula. emp. form.  CH mm emp =13 g 78 g 13 g = 6 C6H6C6H6

A compound has g nitrogen, g oxygen, and molar mass 92 g. Find molecular formula. mm emp =46 g = 2 N2O4N2O4 92 g 46 g NO 2

Island Diagram: a. Diagram has four islands. b. “Mass Island” for elements or compounds c. “Particle Island” for atoms or molecules d. “Volume Island”: for gases only 1 STP = 22.4 L = 22.4 dm 3 1 mol = 6.02 x particles MOLE (mol) Mass (g) Particle (at. or m’c) 1 mol = molar mass (in g) Volume (L or dm 3 ) 1 mol = 22.4 L 1 mol = 22.4 dm 3

1.29 mol What mass is 1.29 mol ? How many molecules is 415 L at STP? sulfur dioxide Fe 2+ NO 3 1– Fe(NO 3 ) 2 () 1 mol g = 232 g sulfur dioxide SO L () 1 mol 22.4 L () 1 mol 6.02 x m’c = 1.12 x m’c Iron (II) nitrate

22.4 L () 1 mol 87.3 L What mass is 6.29 x m’cules ? Al 3+ SO 4 2– Al 2 (SO 4 ) 3 aluminum sulfate = 3580 g () 1 mol g () 1 mol 6.02 x m’c 6.29 x m’c At STP, how many g is 87.3 L of nitrogen gas? N2N2 () 1 mol 28.0 g = 109 g

During a chem. rxn.; atoms are rearranged (NOT created or destroyed!) Chemical equations must be balanced to show the relative amounts of all substances. Balanced means: each side of the equations has the same # of atoms of each element. CH 4 + O 2 —> H 2 O + CO 2 CH 4 + 2O 2 —> 2H 2 O + CO 2 Unbalanced Balanced

Cellulose reacts with oxygen gas to form carbon dioxide gas & liquid water. C 6 H 10 O 5 + 6O 2  6CO 2 + 5H 2 O 6 C6 10 H10 17 O17 Balanced!!!

Nitroglycerin decomposes to form nitrogen gas, oxygen gas, carbon dioxide gas & water vapor 2 C 3 H 5 (NO 3 ) 3  3N 2 + O 2 + 6CO 2 + 5H 2 O 6C6 10H10 6N6 18O19 Not Balanced!

When balancing equations, you may change coefficients as much as you need to, but you may never change subscripts because you can’t change what substances are involved.

H 2 (g) + O 2 (g)  H 2 O (l) 1.Atom Inventory: Reactants Products 2H22H2 2O12O1 2Add coefficients to balance 3. Double check to make sure it is all BALANCED!

Balancing Chemical Equations practice