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Hydrated Ionic Compounds Hydrated Ionic Compounds

Hydrated Ionic Compounds - PowerPoint Presentation

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Uploaded On 2018-03-11

Hydrated Ionic Compounds - PPT Presentation

You have learned how to find the molar mass of a compound with the help of your periodic table And you now know how to find the empirical and molecular formulae of a compound Sometimes though chemists know most of the compound ID: 647331

water mass find compound mass water compound find molecules hydrate formula 159 100 molar compounds sulfate magnesium hydrated called

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Slide1

Hydrated Ionic Compounds

You have learned how to find the molar mass of a compound with the help of your periodic table. And you now know how to find the empirical and molecular formulae of a compound.

Sometimes though, chemists know most of the compound

s formula, but a significant piece of information is missing.

Some ionic compounds crystallize from a water solution with water molecules incorporated into their structure. These are called hydrates.Slide2

Hydrates have a specific number of water molecules chemically bonded to each formula unit.

Epsom salts for example, consist of magnesium sulfate with seven waters attached.

MgSO

4

.

7H

2

O

This is called Magnesium sulfate heptahydrate

Compounds that have no water molecules incorporated into them are called anhydrous. This distinguishes them from their hydrated forms.Slide3

The molar mass of a hydrated compound must include the mass of any water molecules that are attached to the compound.

Magnesium sulfate

heptahydrate

is calculated as follows:

Mg 24.305

S 32.065

4O 4(15.999)

7 H

2

O 7(18.015)

246.471 g/

molSlide4

Determining the Formula of a Hydrate

A hydrate of Barium hydroxide, Ba(OH)

2

.

?H

2

O, is used to make barium salts and to prepare certain organic compounds.

If you have a 50.0 g sample of the hydrate and it contains 27.2 of Ba(OH)

2

, calculate the percent by mass of water in the hydrate.

Then find out how many water molecules are attached to each Barium hydroxide.Slide5

The total mass of the compound is 50.0 g and 27.2 g is from the Ba(OH)

2

.

50.0 – 27.2 = 22.8 g of water in the sample.

To find the percent mass of water, take the mass of water and divide it by the total, then multiply by 100%.

22.8/50.0 x 100% = 45.6%

Or to find percent mass of Ba(OH)

2

take it

s mass, divide by total, and multiply by 100.

27.2/50.0 x 100% = 54.4%Slide6

Find the empirical formula.

moles of Ba(OH)

2

=

mass of Ba(OH)

2

=

27.2

= 0.159

molar mass from PT 171.341

moles of H

2

O =

mass of H

2

O

=

22.8

= 1.27

molar mass from PT 18.015

Ba(OH)

2

 0.159/0.159 = 1

H

2

O  1.27/0.159 = 7.9 ~ 8

Ba(OH)

2

.

8H

2

O