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Empirical Formula of a Hydrate

Empirical Formula of a Hydrate

Assessment

Presentation

Chemistry

10th Grade

Hard

Created by

Joseph Anderson

FREE Resource

5 Slides • 5 Questions

1

Lab: Hydrates

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2

Multiple Choice

What is the percent composition by mass of oxygen in NaOH (gram formula mass = 40 g/mol)


What the identity of the part in the question above?

1

Sodium (Na)

2

Oxygen (O)

3

Hydrogen (H)

3

Multiple Choice

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What is the percent composition by mass of oxygen in NaOH (gram formula mass = 40 g/mol)


What is the mass of the part (Oxygen (O))?

1

1

2

8

3

16

4

23

4

Multiple Choice

What is the percent composition by mass of oxygen in NaOH (gram formula mass = 40 g/mol)


What is the whole in this problem?

1

1

2

16

3

23

4

40

5

Open Ended

Using the formula we learned yesterday,


%comp = (mass of part/mass of whole)*100


Determine the percent composition by mass of oxygen in NaOH (gram formula mass = 40 g/mol)


mass of part: 16

mass of whole: 40

6

Lab: Anhydrous

Today we will be watching a video of a lab in which Hydrated Copper Sulfate (copper sulfate with water) is dried out to produce anhydrous Copper Sulfate (copper sulfate without water)

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7

Data Collection

The lab will provide you with measurements a, b and d.


The remainder of the measurements can either be taken from those sections or calculated.


The bolded indicators to the side will help. For example, to get c, you must subtract a-b. And e is simply the same answer as b.

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8

Multiple Choice

What is the percent composition by mass of water in CuSO4 * 5H2O?


What is the part that we are trying to determine?

1

water

2

CuSO4

3

CuSO4 * 5H2O

9

The Whole

In this case, the whole is not the gram formula mass, as we learned yesterday, but simply the mass of CuSO4 * 5H2O that we started with.


The mass of a.

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10

Final Calculations

Like yesterday,


mass of the part (the mass of water, located at point i) and that is our numerator (top number)


mass of the whole (the mass of the initial compound, located at letter g) and that is the denominator (bottom number)


We then divide the two numbers and multiply by a 100

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Lab: Hydrates

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