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MOLES

- This unit test contains 6 types of problems:
- Molar mass and % composition—must be able to write a chemical formula
- Grams to moles (using molar mass)
- Moles to particles/atoms/molecules (using Avogadro’s #)
- Grams to moles to particles (using molar mass and Avogadro’s #)
- Empirical formulas
- Molecular formulas

Mole

- The SI (metric) unit used to measure the amount of a substance
- 1 mole is always equal to:

--Molar mass (g/mole)

--Avogadro’s number of particles (6.02 x 1023)

--22. 4 Liters of a GAS (AKA molar volume)

These may be used as conversion factors when working mole problems.

activator

- Define molar mass AND Avogadro’s number.

Molar mass of elements—same as atomic mass

- Ex: Titanium 47.867 = 47.9 g/mole

(this sample contains Avogadro’s number of atoms)

- Ex: oxygen 15.999 = 16.0 g/mole

(this sample contains Avogadro’s number of atoms)

Molar Mass of Compounds—must write chemical formula correctly

- Multiply the # of atoms for each element by the atomic mass from periodic table

Ex:Magnesium hydroxide

Mg (OH)2

Mg 1(24.3) = 24.3

O 2(16.0) = 32.0

H 2(1.0) = 2.0

58.3 g/mole (this mass also contains Avogadro’s number of molecules)

Find the molar mass of aluminum sulfate

Al 2 (SO4)3

Al 2(27.0) = 54.0

S 3 (32.1)= 96.3

O 12(16.0) = 192.0

342.3 g/mole (this mass also contains Avogadro’s number of molecules)

MOLAR MASS PRACTICE--

- Stannic carbonate
- Diarsenicpentasulfide
- Hydrofluoric acid

% composition

- Shows the % of each element that makes up a compound
- Must be calculate molar mass first.

Ex: magnesium hydroxide Mg (OH)2

Mg 1 x 24.3 =24.3

24.3/58.3 x 100 = 41.7%

O 2 x 16= 32.0

32.0/58.3 x 100 = 54.9%

H 2x1.0 = 2.0 58. 3 g/mole

2.0/58.3 x 100 = 3.4%

ACTIVATOR:

- Calculate the % composition of sulfurous acid

LAB—BITE THE BUBBLE

- SAVE YOUR WRAPPER FOR ENTIRE LAB!!
- DO NOT START CHEWING UNTIL YOU SIT DOWN BACK AT YOUR DESK.
- CHECK BALANCE TO MAKE SURE IT’S OK BEFORE YOU START!!

READ PROBLEM:

- MAKE A HYOTHESIS:
- PROCEDURE 1—4
- DATA TABLE 1—5

After chewing: (KEEP YOUR SAME BALANCE)

- Procedure 5—8
- Data table 6—8
- Conclusion
- Questions 1—2

C 12( 12.0) = 144.0

- H 19 (1.0) = 19.0
- Cl 3 (35.5) = 106.5
- O 8 (16.0) = 128.0

397. 5 g/mole

% C= 144.0 / 397.5 x 100 = 36.2%

% H= 19.0 / 397.5 x 100 = 4.8%

% Cl= 106.5 / 397.5 x 100 = 26.8%

% O= 128.0/ 397.5 x 100 = 32.2%

Ticket out

- 1. Calculate the % composition of carbonic acid.
- 2. Calculate the % composition of diantimony trioxide.

1. H2CO3

H-2 (1.0) = 2.0 3.2%

C- 1 (12.0) = 12.0 19.4%

O – 3(16.0) = 48.0 77.4%

62.0 g/mole

- Sb2O3

Sb- 2(121.8) =243.6 83.5%

O – 3(16.0) = 48.0 16.5%

291.6 g/mole

ACTIVATOR

- Calculate the molar mass AND % composition of:
- C12H22O11
- Cupric sulfate

C: 12 (12.0) = 144.0

- H: 22 (1.0) = 22.0
- O: 11(16.0) = 176.0

CuSO4

Cu: 1(63.5)= 63.5

S: 1 (32.1) = 32.1

O: 4(16.0) = 64.0

Converting Grams to moles

Will need to use unit conversion(cancellation) and molar mass will be used for the conversion factor.

Ex: 2.50 grams of hydrochloric acid = ____moles

H Cl

2.50 grams x 1 mole = 0.0685 moles(3sigfigs)

36.5 grams

Converting moles to grams

Ex: 2.50 moles of HCl = __________grams

2.5 moles x 36.5 grams = 91 grams

1 mole (2 sig figs)

Converting particles to moles

- Particles, atoms, molecules (synonyms)
- Will have to use Avogadro’s number as a conversion factor
- Ex: 5.25 x 1025 atoms of Mg = _____moles

5. 25 x 1025 atoms x 1 mole = 87.2 moles

6.02 x 1023 (3 sig figs)

Converting moles to particles

2.50 moles MgO = _________molecules

2.50 moles x 6.02 x 1023 molecules

1 mole

= 1.50 x 1024 molecules (3 sig figs)

Converting grams to particles

- Will need to use both molar mass AND Avogadro’s number as conversion factors
- Will be 2 steps instead of 1 step unit cancellation
- Ex: 4.5 grams nitrous acid = __________molecules
- HNO2

4.5 g x 1 mole x 6.02 x 1023 molecules =

47 g 1 mole

5.8 x 1022 molecules (2 sig figs)

Converting particles to grams

- Ex: 9.35 x 1021 particles of carbon tetrabromide = _____grams

C Br4

9.35 x 1021 p x 1 mole x 154 grams =

6.02x1023 p 1 mole

2.39 grams (3 sig figs)

Answers to homework: must show work to receive credit

1. 0.14 mole (gram to moles)

2. 150 g (moles to grams)

3. 1.1 x 1023 molecules (g to molecules)

4 5.30 x 1025 molecules (moles to molecules)

5. 0.074 mole( gram to moles)

6. 0.619 g (particles to grams)

7. 0.49 mole (grams to moles)

8. 0.00083 mole (particles to moles)

Tab 3 “even more mole problems”—show work to receive credit

- 22 g
- 1.53 x 1024 molecules
- 0.014 g
- 7.2 x 1021 molecules
- 7.2 x 1023 molecules
- 2.08 x 106 g
- 56 g
- 2.5 g
- 31 g
- 0.0029 mole
- 167 g

BONUS: 3.37 x 1026 atoms

activator

If grams are converted to moles,

use _______________ to convert.

If moles are converted to

molecules, then use

______________to convert.

****Have calculator, periodic table, and best friend chart****

Mole game

All members of your group must show their work on separate sheet of paper.

When you calculate the answer, flip the card over to find a word.

All of your words will make a sentence.

First group to show all work and finish first, wins bonus!

Mole problem diagram

Grams-----moles-----particles(atoms or molecules)

Extra practice-mole problems

- 55.33 grams of sodium oxide = ____moles
- 5.00 x 1022 particles of sodium= _______moles
- 2.49 x 1026 atoms of acetic acid = _______________grams

Homework:

Video sheet Problems #1—3, 5, 6

ACTIVATOR:

Fill in the blanks with multiply/divide OR molar mass/Avogadro’s number:

***When going from moles to grams, _______________ by _____________.

***When going from moles to particles, ____________by _________________.

I AM CHECKING 5 HOMEWORK PROBLEMS!!!!

Ticket out

Briefly describe the steps for calculating an empirical formula AND molecular formula.

Mole review problems

- Convert 5.03 x 1024 molecules of phosphoric acid to grams.
- Convert 35.75 grams of dinitrogen monoxide to moles.
- Convert 5.0 moles of water to molecules.

answers

- 819 grams of H3PO4
- 0.8125 moles of N20
- 3.0 x 1024 molecules of water

Ticket out

Tell how to solve for each:

- G to moles
- Moles to G
- Particles to moles
- Moles to particles
- G to particles
- Particles to G

Gum lab—show work to receive credit

- HYPOTHESIS, DATA 1—8, CONCLUSION, QUESTIONS 1—4
- 3. MASS OF SUGAR

(in grams—data #8)--------MOLES

(SUGAR = C12 H22 O11)

- 4. MOLES----------PARTICLES

Lab-using the mole

- Data Table: mass of empty vial AND substances mass (make sure you’ve subtracted empty vial each time!!)
- SHOW WORK TO GET CREDIT
- CONVERT GRAMS----------MOLES
- CONVERT MOLES---------PARTICLES
- ANSWER QUESTIONS 1---5, 6 (BONUS)

Activator:

Convert 25.0 moles of water to grams.

activator

A molecular formula is a whole number____________of the empirical formula.

EMPIRICAL FORMULAS

- Shows the SIMPLEST, WHOLE NUMBER ratio of elements in a compound
- Will give you % composition of compound and ask you to find the formulas

Steps for calculating:

1. Change % sign to grams (some problems may already give you grams instead of %)

2. Convert grams to moles (using molar mass)

**round to 4 decimals***

3. Simplify the mole ratio by dividing each one by the smallest

4. Round to the nearest whole number and assign these numbers to the appropriate element

Example:

A compound is 78.1% Boron and 21.9% H. Calculate the empirical formula.

78.1 grams B x 1 mole = 7.2315 moles B

10.8 g

21.9 grams H x 1mole =21.9 moles H

1.0 g

7.2315 : 21.9

7.2315 7.2315

1: 3 = BH3

Molecular formulas:

Is a WHOLE NUMBER MULTIPLE of the empirical formula

You must then first know the empirical formula

Steps for calculating:

- You must first have empirical formula (if not, you will have to calculate it first!!)
- Find the molar mass of the empirical formula.
- Take the molar mass of the molecular formula that is given in the problem divided by the empirical formula’s molar mass. Round this to a whole number.
- Distribute this number to the numbers within the empirical formula to get the new molecular formula.

Example: 1

Given the empirical formula of BH3 and the molecular formula’s molar mass of 27.67 g/mole, find the molecular formula.

Molar mass of BH3 is 13.8 g/mole

27.67 divided by 13.8 = 2

So molecular formula is B2H6

Example: 2

A compound is 4.04 grams of nitrogen and 11.46 grams of oxygen. The molecular molar mass is 108.0 g/mole. Find the molecular formula.

4.04 g N x 1 mole= 0.2886 mole N

14.0 g

11.46 g O x 1 mole =0.7163 mole O

16.0 g

0.2886 : 0.7163

0.2886 0.2886

1: 2 = NO2 empirical molar mass = 46.0 g/mole

108.0 divided by 46.0 = 2

Molecular formula = N2O4

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