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Entry Task: Feb 5 th -6 th Block #1. Write the Question How many formula units are in 50.0 grams of Iron II oxide? You have 5 minutes!. Agenda:. Discuss Particles <-> Moles<-> Grams Self Check on Concepts Discuss Self-Check Notes on Percent Composition HW: Percent Composition ws.

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Entry Task: Feb 5 th -6 th Block #1

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Entry task feb 5 th 6 th block 1

Entry Task: Feb 5th-6thBlock #1

Write the Question

How many formula units are in 50.0 grams of Iron II oxide?

You have 5 minutes!


Entry task feb 5 th 6 th block 1

Agenda:

Discuss Particles <-> Moles<-> Grams

Self Check on Concepts

Discuss Self-Check

Notes on Percent Composition

HW: Percent Composition ws


I can

I can…

  • Calculate particles moles  grams

  • Explain what is meant by the percent composition of a compound.


Entry task feb 5 th 6 th block 1

Discuss

Particles  Moles  Grams

Avervision for answers


Entry task feb 5 th 6 th block 1

Clear off your Desk!!

Calculators

Periodic Tables

Needed


Entry task feb 5 th 6 th block 1

Discuss

Self Check

Avervision for answers


Entry task feb 5 th 6 th block 1

Percent Composition is the percent of each element in a compound.

Percent composition is a step in determining the ratio in elements in a compound/molecule.


Entry task feb 5 th 6 th block 1

Example: Magnesium oxide- MgO

By just looking at this compound, It LOOKS like it 50/50 of magnesium oxide.

But according to its molar mass, it is not 50/50.

Its molar mass is

24.305 g for Mg + 15.999 g for O, = 40.304 g

24.305g (Magnesium mass) 40.304 grams (mass of MgO)

Magnesium

= 0.60 X 100 = 60%

Oxygen

15.999 g (oxygen mass)

40.304 grams (mass of MgO)

= 0.40 X 100 = 40%


Entry task feb 5 th 6 th block 1

Iron III oxide

Fe2O3

Fe = 55.845 x 2 = 111.69 g

 159.687 g

0.6994 X 100 = 69.9%

 159.687 g

O = 15.999x 3 = 47.997g

0.3005 X 100 = 30.1%

159.687 g


Entry task feb 5 th 6 th block 1

Barium hydroxide Ba(OH)2

Ba= 137.33 g X 1 = 137.33 g

 171.3438 g

= 0.801 X 100 = 80.1%

O= 15.999 g X 2 = 31.998 g

 171.3438 g

= 0.187 X 100 = 18.7%

H= 1.0079 g X 2 = 2.0158 g

 171.3438 g

= 0.0117 X 100 = 1.2%

171.3438 g


Entry task feb 5 th 6 th block 1

Lead (II) nitrate Pb(NO3)2

Pb= 207.2 g X 1 = 207.2 g

 331.208 g

= 0.626 x100 = 62.6%

N= 14.007 g X 2 = 28.014 g

 331.208 g

= 0.0845 x100 = 8.46%

 331.208 g

= 0.289 x100 = 28.9%

O= 15.999 g X 6 = 95.994 g

331.208 g


Entry task feb 5 th 6 th block 1

Ammonium sulfate (NH4)2SO4

 131.044 X 100 = 21.4 %

N= 14.007 g X 2 = 28.014 g

H= 1.0079 g X 8 = 8.06 g

 131.044 X 100 = 6.2%

S= 30.974 g X 1 = 30.974 g

 131.044 X 100 = 23.6 %

O= 15.999 g X 4 = 63.996 g

 131.044 X 100 = 48.8.4 %

add up % = 100%

1 mole = 131.044 grams


Entry task feb 5 th 6 th block 1

Hydronium nitride

(H3O)3N

 71.1 grams x 100=

H= 1.0079 g X 9 = 9.071 g

= 12.7 %

O= 15.999 g X 3 = 47.997 g

 71.1 grams x 100=

= 67.6 %

N= 14.007 g X 1 = 14.007 g

 71.1 grams x 100=

= 19.7 %

71.1 grams


Entry task feb 5 th 6 th block 1

Aluminum cyanide

Al(CN)3

Al= 26.98 g X 1 = 26.98 g

 105 grams x 100=

= 25.6 %

 105 grams x 100=

= 34.3 %

C= 12.0107 g X 3 = 36.032 g

 105 grams x 100=

= 40.0 %

N= 14.007 g X 3 = 42.021 g

105 grams


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