+ All Categories
Home > Documents > Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Date post: 16-Jan-2016
Category:
Upload: damon-webster
View: 215 times
Download: 0 times
Share this document with a friend
Popular Tags:
34
Review for Review for Tomorrow’s Test Tomorrow’s Test Moles Moles Stoichiometry Stoichiometry Percent composition Percent composition Empirical Formula Empirical Formula
Transcript
Page 1: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Review for Tomorrow’s TestReview for Tomorrow’s Test

MolesMolesStoichiometryStoichiometry

Percent composition Percent composition Empirical FormulaEmpirical Formula

Page 2: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Who is this famous Italian scientist,and what does his number represent?

Page 3: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

He is Amadeo Avogadro His number: 6.022 x 1023

There are 6.022 x 1023

molecules of any substance in a mole of that substance

Page 4: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Define what a mole is:

A mole of any substance is __ _________

______ _________ __ _____.

Page 5: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

A mole of any substance is its molecular weight expressed in grams

Page 6: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How much does one mole of H2SO4 weigh?

Page 7: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

H2SO4

H 2 x 1.0 = 2.0S 1 x 32.1 = 32.1O 4 x 16.0 = 64.0 98.1 grams/mole

Page 8: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many moles of NaCl are there in 14.65 g ?

Page 9: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Number of grams/mole: Na 1 x 23.0 = 23.0 Cl 1 x 35.5 = 35.5

58.5 g/moleNumber of moles/ liter: 14.65 g x 1 mole = 0.25 moles 58.5 g

Page 10: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Write a balanced equation for the formation of salt

(sodium chloride) from sodium metal and chlorine gas

Page 11: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

2 Na (s) + Cl2 (g) 2 NaCl (s)

(This is a combination or synthesis reaction)

Page 12: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many grams are in 19.3 moles of AlCl3?

Page 13: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many grams are in 19.3 moles of AlCl3?

19.3 moles x 133.5 g = 2,576.6 g of AlCl3

1 mole

Page 14: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many moles are in 355.3 grams of Na2SO4?

Page 15: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many moles are in 355.3 grams of Na2SO4?

355.3 g x 1 moles = 2.5 moles of Na2SO4

142.1 g

Page 16: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many grams does 1.807 x 1024 molecules of water weigh?

Page 17: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many grams does 1.807 x 1024 molecules of water weigh?

1.807 x 1024 molecules x 1 mole x 18.0g = 54.0 g 6.02 x 1023molecules 1 mole

Page 18: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many molecules are in 112 grams of CH4?

Page 19: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

How many molecules are in 112 grams of CH4?

112 g x 1 mole x 6.02 x 1023 molecules = 4.214 x 1024

16.0 g 1 mole molecules

Page 20: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Determine the percent composition of Determine the percent composition of

Each element in HEach element in H22SOSO44

Page 21: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Determine the percent composition of Determine the percent composition of

Each element in HEach element in H22SOSO44

H = 2 x 1.0 = 2.0 gS = 1 x 32.1 = 32.1 gO = 4 x 16.0 = 64.0 g

98.1 g/mole

H = 2.0g x 100 = 2.0% 98.1gS = 32.1g x 100 = 32.7% 98.1gO = 64.0g x 100 = 65.2% 98.1g

Page 22: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

An unknown compound is found to be An unknown compound is found to be 75% Carbon and 25% Hydrogen 75% Carbon and 25% Hydrogen Determine it’s Empirical FormulaDetermine it’s Empirical Formula(Use a 100 g sample )(Use a 100 g sample )

Page 23: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

An unknown compound is found to be An unknown compound is found to be 75% Carbon and 25% Hydrogen 75% Carbon and 25% Hydrogen Determine it’s Empirical FormulaDetermine it’s Empirical Formula(Use a 100 g sample )(Use a 100 g sample )

If there is a 100.0 g sample, then 75.0 g of it is C and 25.0 g of it is H

Carbon: 75.0 g x 1 mole = 6.25 moles 12.0 g

Hydrogen: 25.0 g x 1 mole = 25.0 moles 1.0 g

Find the ratio:6.25 = 16.25

25.0 = 46.25

CH4

Page 24: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

HCl + NaHCl + Na22COCO33 CO CO22 + H + H22O + NaClO + NaCl

How many grams of HCl and NaHow many grams of HCl and Na22COCO33 would would

you need to make 35.1 g of salt (NaCl)?you need to make 35.1 g of salt (NaCl)?

35.1 g = 0.6 moles of NaCl. Therefore 0.6 mole of HCl and

0.3 moles of Na2CO3 are needed

HCl needed: 0.6 moles x 36.5 g/mole = 21.9 g

Na2CO3 needed: 0.3 moles x 106 g/mole = 31.8 g

Page 25: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Balance this equation:

Al (s) + O2 (g) Al2O3 (s)

Page 26: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Balanced !

4Al (s) + 3 O2 (g) 2Al2O3 (s)

Page 27: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

__Ca + __O2 __CaO

Balance the equation and determine the Molar Ratio needed

between calcium and oxygen?

Page 28: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

2Ca + O2 2CaOA 2 to 1 ratio between calcium

& oxygen

Page 29: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Describe what is meant by a

polyatomic ion

Page 30: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

A group of atoms that bind together

sharing an ionic charge and acting as one.

Page 31: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

What is the sulfate radical and its valence charge?

Page 32: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Sulfate = SO4-2

Page 33: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

What is a phosphate polyatomic ion and

what is its valance chargewhen it forms a compound?

Page 34: Review for Tomorrow’s Test MolesStoichiometry Percent composition Empirical Formula.

Phosphate = PO4-3


Recommended