convert the following:convert the following:1.1. Find the % composition of AlFind the % composition of Al

(NO(NO33))33

2.2. If 8.52 grams of aluminum If 8.52 grams of aluminum nitrate are dissolved in nitrate are dissolved in exactly 250 cmexactly 250 cm33, what is the , what is the molarity?molarity?

3.3. ACT-prep question:ACT-prep question: Which chart would you use to Which chart would you use to

answer the following answer the following question?question?

Do the data support the Do the data support the hypothesis that frogs in hypothesis that frogs in warmer climates will be warmer climates will be noisier than frogs in colder noisier than frogs in colder climates?:climates?:

Quiz:Quiz: March 15, 2005

1. Al = 27 amu, N = 14 amu 0 = 16 amu

(27.0) + 3(14.0) + 9(16) =213 amuor 213 g/mole27/213 x 100% = 12.7% Al42/213 x 100% = 19.7 % N144/213 x100% = 67.6 % O

2. 8.52 gAl(NO3)3 mole_Al(NO3)3_1000 cm3

250 cm3 213 g Al(NO3)3 1 dm3

0.160 M Al(NO3)3

Time of Day Time of Day (P.M.)(P.M.)

Total number of Total number of predatorspredators

A.A.

Population Population sizesize

Average Average call ratecall rate

Average Average Call Call volumevolume

C.C.

Time of Time of DayDay

(P.M.)(P.M.)

Average Average call ratecall rate

Average Average Call Call volumevolume

B.B.

Water Water TemperaturTemperaturee

Average Average call ratecall rate

Average Average Call Call volumevolume

D.D. TestThis

Thursday

HW 18: #56 HW 18: #56 64 p210; Honor’s 64 p210; Honor’s #103 a,b and 104 a, b p 218#103 a,b and 104 a, b p 218

Empirical Formula, first introduced in Chapter 7, refers to the experimental data you obtain when trying to find out how much of each element is in a sample of a compound.

Basically, you do the opposite of the procedure to find % composition. Use grams to find a ratio of part:whole, convert to moles, find a mole ratio. Write an empirical formula

.57 g of Magnesium burns in air. 0.96 g .57 g of Magnesium burns in air. 0.96 g of magnesium oxide is measured after of magnesium oxide is measured after combustion is complete. What is the combustion is complete. What is the empirical formula of magnesium oxide?empirical formula of magnesium oxide?

Underlineimportant

information

List whatyou have,and whatyou want.

0.57 g Mg0.96 g Mg?O?

____g Ofind mole ratioMg: O

0.57 g Mg0.96 g Mg?O?

0.39 g O

Subtract0.96 – 0.57

to findg O

Find the# molesof eachelement

0.57 g Mg24.3 g/mole Mg

0.39 g O16.0 g/moleO

0.0235 mole Mg

0.0244moles O

0.0235

0.0235

= 1 mole Mg

= 1 mole O

MgO

A compound has a percentage composition A compound has a percentage composition of 40% C, 6.71 % H, and 53.3% Oof 40% C, 6.71 % H, and 53.3% O

40 g C40 g C

12 g/mole C12 g/mole C

6.71 g H6.71 g H

1.01 g/mole C1.01 g/mole C

53.3 g O53.3 g O

16.0 g/mole O16.0 g/mole O

3.33 mole C3.33 mole C

6.64 mole H6.64 mole H

3.33 mole O3.33 mole O

____________3.33

____________3.33

____________3.33

1 mole C1 mole C

2 mole H2 mole H

1 mole O1 mole O

CH2OStep 1: assume % = gramsStep 2: change grams molesStep 3: Find whole number ratio Step 4: Use ratio to write formula

Analysis of a Analysis of a 10.150 g 10.150 g sample of a compound sample of a compound known to contain known to contain only phosphorous and oxygenonly phosphorous and oxygen yields yields 5.717 g of oxygen5.717 g of oxygen. What is the empirical . What is the empirical

formulaformula of this compound? of this compound? 10.150 – 5.71710.150 – 5.717

= 4.433 g P= 4.433 g P

4.433 g P4.433 g P

30.97 g/mole P30.97 g/mole P

5.717 g O5.717 g O

16.00 g/mole O16.00 g/mole O

0.1431 mole 0.1431 mole PP

0.3573 mole 0.3573 mole OO

____________0.1431

____________0.1431

1 mole P1 mole P

2.497 mole O2.497 mole O

= 2 ½ mole O= 2 ½ mole O

P2 O 2(2.5)

Step 1:get grams of each thing.Step 2: change grams molesStep 3: Find whole number ratio Step 4: Use ratio to write formula

P2O5

A compound has a percentage composition of 40% C, A compound has a percentage composition of 40% C, 6.71 % H, and 53.3% O, for the following molecular 6.71 % H, and 53.3% O, for the following molecular

masses, what would the actual formula be?masses, what would the actual formula be?40 g C40 g C

12 g/mole C12 g/mole C

6.71 g H6.71 g H

1.01 g/mole C1.01 g/mole C

53.3 g O53.3 g O

16.0 g/mole O16.0 g/mole O

3.33 mole C3.33 mole C

6.64 mole H6.64 mole H

3.33 mole O3.33 mole O

____________3.33

____________3.33

____________3.33

1 mole C1 mole C

2 mole H2 mole H

1 mole O1 mole O

CH2O

a.30 g/moleb.60 g/molec.120 g/moled.240 g/mole

C = 12C = 12 2H = 22H = 2 O = O = 16_________16_________ CHCH22O = 30 g/moleO = 30 g/mole

a.a. 30/ 30 = 1 CH30/ 30 = 1 CH22OO

b.b. 60/30 = 2 C60/30 = 2 C22HH44OO22

c.c. 120/30 =4 C120/30 =4 C44HH88OO44

d.d. 240/30 = 8 C240/30 = 8 C88HH1616OO88

% water in a hydrate% water in a hydrate We have a 10.407 g sample of We have a 10.407 g sample of

hydrated barium iodide. The sample hydrated barium iodide. The sample is heated to drive off the water. The is heated to drive off the water. The dry sample has a mass of 9.520 g. dry sample has a mass of 9.520 g. What is the formula of the hydrate?What is the formula of the hydrate?

10.407 g hydrated BaI2

9.520 g BaI2

g water driven off0.887

Ba = 1372 I = 2(127) =254 391g/mole2H = 2(1.01)O = 16.0 18.0 g/mole water

9.520 g BaI2 = 0.0243 mole BaI2

391g/mole0.887 g water = 0.493 mole H2O 18.0 g/mole water

0.0243

0.0243

1 BaI2 : 2 H2Oso…

BaI2. 2 H2O