experiment 11
A. Finding the Simplest Formula
The simplest formula of a compound represents the lowest whole-number ratio of the atoms in that compound.
Example of Compounds and Their Formulas:
Benzene:
Molecular formula: C_6H_6
Simplest formula: CH
Others:
Compound Name
Molecular Formula
Simplest Formula
Acetylene
C_2H_2
CH
Benzene
C_6H_6
CH
Ammonia
NH_3
NH_3
Hydrazine
N_2H_4
NH_2
To determine the simplest formula:
Convert the grams of each element to moles.
Find the lowest whole-number ratio to use as subscripts in the formula.
Example Calculation:
Given:
0.040 moles of Zn
0.080 moles of Cl
Step 1: Divide each element's moles by the smaller number of moles:
rac{0.080 ext{ mole Cl}}{0.040} = 2 ext{ moles of Cl}
rac{0.040 ext{ mole Zn}}{0.040} = 1 ext{ mole of Zn}
Step 2: Use the whole numbers as subscripts to write the formula:
ZnCl_2 (the subscript 1 for Zn is understood).
B. Formula of a Hydrate
Definition: A hydrate is an ionic compound that is combined with a fixed specific number of water molecules.
This number of water molecules may differ from one hydrate to another.
The format to express the number of water molecules is as follows:
Written after the ionic formula and separated by a raised dot (•).
Examples of Hydrate Formulas:
Calcium Sulfate Dihydrate: CaSO_4 ullet 2H_2O
Copper(II) Sulfate Pentahydrate: CuSO_4 ullet 5H_2O
Sodium Carbonate Decahydrate: Na_2CO_3 ullet 10H_2O
Removal of Water:
Water molecules in a hydrate can be removed by heating.
After heating, the remaining ionic compound is termed an anhydrate.
Example reaction:
When heating Copper(II) Sulfate Pentahydrate: CuSO_4 ullet 5H_2O ightarrow CuSO_4 + 5H_2O(g)
The removed water is referred to as water of hydration.
Experimentally Determining the Amount of Water:
Measure the mass of the hydrate before heating and the mass of the anhydrate after heating.
The difference in mass reflects the amount of water removed, which is the water of hydration.
Example Calculation of Percent Water in a Hydrate:
Given a sample of 2.00 ext{ g} of CuSO_4 ullet 5H_2O heated to produce 1.28 ext{ g} of anhydrate CuSO_4:
Calculate grams of water lost:
2.00 ext{ g of hydrate} - 1.28 ext{ g of anhydrate} = 0.72 ext{ g of } H_2O ext{ in hydrate}
Calculate percent of water in the hydrate:
rac{0.72 ext{ g } H_2O}{2.00 ext{ g } CuSO_4 ullet 5H_2O} imes 100 ext{%} = 36 ext{% } H_2O ext{ in hydrate}