Unit 6: The Mathematics of Chemical Formulas Chemistry Na 2 CO H 2 O p v MOL m Cu(OH) 2 SO 3

# of H 2 O molecules # of H atoms# of O atoms x g (6.02 x )6.02 x g 18.0 g molar mass: the mass of one mole of a substance

PbO 2 HNO 3 ammonium phosphate 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 (NH 4 ) 3 PO 4 H: 12 (1.0 g) = 12.0 g N: 3 (14.0 g) = 42.0 g g P: 1 (31.0 g) = 31.0 g NH 4 1+ PO 4 3– O: 4 (16.0 g) = 64.0 g

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)

zinc acetate Zn 2+ CH 3 COO 1– Zn(CH 3 COO) g = 3.3% H = 34.9% O = 35.7% Zn = 26.2% C : C: 4 (12.0 g) = 48.0 g Zn: 1 (65.4 g) = 65.4 g H: 6 (1.0 g) = 6.0 g O: 4 (16.0 g) = 64.0 g g

“What’s your flavor of ice cream?” 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

A ruthenium/sulfur compound is 67.7% Ru. Find its empirical formula. RuS 1.5 Ru 2 S 3

A g sample of a technetium/oxygen compound contains g of Tc. Find the empirical formula. TcO 3.5 Tc 2 O 7

A compound contains 4.63 g lead, 1.25 g nitrogen, and 2.87 g oxygen. Name the compound. PbN 4 O 8 Pb(NO 2 ) 4 Pb ? 4 NO 2 1– lead (IV) nitrite (plumbic nitrite) ? ?

(How many empiricals “fit into” the molecular?) To find molecular formula… A. Find empirical formula. B. Find molar mass of empirical formula. C. Find n = mm molecular mm empirical D. Multiply all parts of empirical formula by n. (“How many scoops?”) (“What’s your flavor?”)

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

Mole Calculations 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

New Points about Island Diagram: a. Diagram now has four islands. b. “Mass Island” now for elements or compounds c. “Particle Island” now 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 ferrous nitrate () 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

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 dm 3 of nitrogen gas? N2N2 () 1 mol 28.0 g = 109 g

But there are 9 atoms per molecule, so… How many m’cules is 315 g of iron (III) hydroxide? Fe 3+ OH 1– Fe(OH) g () 1 mol g () 1 mol 6.02 x m’c = 1.78 x m’c How many atoms are in 145 L of CH 3 CH 2 OH at STP? 145 L () 1 mol 22.4 L () 1 mol 6.02 x m’c = 3.90 x m’c 9 (3.90 x ) = 3.51 x atoms

Hydrates and Anhydrous Salts anhydrous salt: an ionic compound (i.e., a salt) that attracts water molecules and forms loose chemical bonds with them; symbolized by MN “anhydrous” = “without water” Uses: hydrate: an anhydrous salt with the water attached -- symbolized by MN. ? H 2 O Examples: “desiccants” in leather goods, electronics, vitamins CuSO 4. 5 H 2 O BaCl 2. 2 H 2 O Na 2 CO H 2 O FeCl 3. 6 H 2 O

ENERGY + + MN H2OH2O H2OH2O H2OH2O H2OH2O H2OH2OH2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O H2OH2O HEAT + hydrate anhydrous salt water

Finding the Formula of a Hydrate 1. Find the # of g of MN and # of g of H 2 O. 2. Convert g to mol. 3. Divide each “# of mol” by the smallest “# of mol.” 4. Use the ratio to find the hydrate’s formula.

Find formula of hydrate for each problem. sample’s mass before heating = 4.38 g sample’s mass after heating = 1.93 g molar mass of anhydrous salt = 85 g MN. ? H 2 O MN (hydrate) (anhydrous salt) MN. 6 H 2 O

beaker + salt + water A. beaker = g B. beaker + sample before heating = g C. beaker + sample after heating = g molar mass of anhydrous salt = g beaker + salt MN. 8 H 2 O

beaker + salt + water A. beaker = g B. beaker + sample before heating = g C. beaker + sample after heating = g molar mass of anhydrous salt = 128 g beaker + salt MN. 4 H 2 O

or… For previous problem, find % water and % anhydrous salt (by mass).

Review Problems Find % comp. of Fe 3+ Cl 1– FeCl 3 iron (III) chloride. Fe: 1 (55.8 g) = 55.8 g Cl: 3 (35.5 g) = g g  34.4% Fe  65.6% Cl :

A compound contains g C and g H. Its molar mass is 58 g. Find its molecular formula. emp. form.  C 2 H 5 mm emp =29 g 58 g 29 g = 2 C 4 H 10

At STP, how many g is 548 L of chlorine gas? 22.4 L 548 L () 1 mol Cl 2 () 1 mol 71.0 g = 1740 g

beaker + salt + water A. beaker = 65.2 g B. beaker + sample before heating = g C. beaker + sample after heating = g beaker + salt SrCl 2. 6 H 2 O is an anhydrous salt on which the following data were collected. Find formula of hydrate. Strontium chloride Sr 2+ Cl 1– SrCl 2

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