convert the following:convert the following:1.1. (4.0 x 10(4.0 x 1066) X 0.0040 = _____) X 0.0040 = _____2.2. Find the gfw of AlFind the gfw of Al (NO(NO33))333.3. How many moles are in 0.0426 How many moles are in 0.0426

grams of aluminum nitrate?grams of aluminum nitrate?4.4. If 0.852 grams of aluminum If 0.852 grams of aluminum

nitrate are dissolved in exactly nitrate are dissolved in exactly 500 cm500 cm33, what is the molarity?, what is the molarity?

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

answer the following question?answer the following question?Do the data support the hypothesis Do the data support the hypothesis

that frogs in small populations that frogs in small populations call more frequently than frogs call more frequently than frogs in large populations:in large populations:

Quiz:Quiz: March 14, 2005

1. 1.6 x 104

2. Al = 27 amu, N = 14 amu 0 = 16 amu(27) + 3(14) + 9(16) =213 amuor 213 g/mole3. 0.0426 g mole_Al(NO3)3

213 g2.00 x 10-4 mole aluminum nitrate

4. 0.852 gAl(NO3)3 mole_Al(NO3)3_1000 cm3

500 cm3 213 g Al(NO3)3 1 dm3

0.00800 M Al(NO3)3

note, .994 ~ 1Time 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.

Moles

mass ingrams[concentration]

? M

x GFW dm3

X dm3 GFW

Find the molarity of the following solutions:

Concentration Moles cubic centimeters gramsname of

compound Formula

1  .042 M .21 mole  5,0000dm3 12g sodium chloride NaCl 

2

0.10001.0000 

= .1000M0.1000mol

e 1,000. dm3

(.10mole)(56g) (mole)

= 5.6 g  

calcium oxide

40 + 16 = 56 g/mole  CaO

3 0.1M0.1x 0.05

= .005mole  0.05000dm3

.005 mole ( 133 g)  mole

= 0.7 g

copper (II) chloride

63 + 2(35) = 133g/mole CuCl2 

4 0.02M  0.2x 0.05= .01mole 0.05000dm3

.01mole ( 169 g)  mole= 2 g

silver nitrate107 + 14 +

3(16)169 g/mole AgNO3

5 0.666M 3.00mole

3.00 mole0.666

mole/dm3 = 4.50 dm3 =

4500 cm3

3.00 mole ( 169 g)  mole

= 507 g  

silver nitrate107 + 14 +

3(16)169 g/mole AgNO3  

6 0.0033mole0.333 dm3

0.0099 M .0033mole 0.333.00dm3

.0033mole ( 56 g)  mole

= 0.18 g  Sodium chloride  NaCl

7 0.50 M .0010mole

 0.50 mole0.0010

mole/dm3 = 500 dm3 =5 x 10 5 cm3

.0010mole ( 124 g)  mole

= 0.18 g  

copper (II) carbonate64 + 12 +

3(16)= 124 g/mole CuCO3

To Find % compositionTo Find % compositionfind gfwfind gfw

Divide mass of each partDivide mass of each partby the whole gfw.by the whole gfw.

CaCa33(PO(PO44))223 Ca 3(40.1)3 Ca 3(40.1)2 P 2(31.0)2 P 2(31.0)8 O 8(16.0)8 O 8(16.0)

= 120.= 120.= 62.0= 62.0= 128= 128

3 Ca 120.3 Ca 120.2 P 62.02 P 62.08 O 8 O 128 128 310 g/mole310 g/mole

3 Ca 120.g/310.g (100%) = 38.7%3 Ca 120.g/310.g (100%) = 38.7%2 P 62.0g/ 310g (100%) = 20.0%2 P 62.0g/ 310g (100%) = 20.0%8 O 128g/310 g (100%) = 41.3 %8 O 128g/310 g (100%) = 41.3 % The sum

should bewithin 1%of 100%

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

HW 17: #50-55 page 208; HW 17: #50-55 page 208; #101p218#101p218

% = the proportion out of 100 pieces% = the proportion out of 100 pieces If there were 100, how much would If there were 100, how much would

be …be … In a package of M&Ms 15 were brown, 6 In a package of M&Ms 15 were brown, 6

were yellow, and 4 were blue. What % were yellow, and 4 were blue. What % were brown?were brown?

15 + 6 + 4 = total number of M&Ms = 2515 + 6 + 4 = total number of M&Ms = 25 15/25 = the fraction which are brown.15/25 = the fraction which are brown. 60% are brown60% are brown 15 x 100%

25

Find the % nitrogen in Find the % nitrogen in ammonia.ammonia.

NHNH33 = ammonia = ammonia

http://www.ausetute.com.au/percentc.html Tutorial for extra help

N = 14.03H = 3 x 1.0 17.0 amu

14.0 amu N

17.0 amu NH3

100%

= 82.4% N inNH3

To find the % composition of HTo find the % composition of H22OO

H = 2(1.01) H = 2(1.01) O = 16.0O = 16.0 HH22O = 18.0 g/moleO = 18.0 g/mole

% H = 0.112 x 100%% H = 0.112 x 100%11.2% H11.2% H

% O = 0.889 x 100%% O = 0.889 x 100%88.9% O88.9% O

%H = %H = 2.02 g H2.02 g H22

18.0 g/mole H18.0 g/mole H22O O

%O = %O = 16.0 g O16.0 g O 18.0 g/mole H18.0 g/mole H22O O