Lab 4C: Formula of a hydrate

Objectives:
1.to determine the percentage of water in an unknown hydrate
2. to determine the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt
3. to write the empirical formula of the hydrate.
Table1
Mass of empty crucible: 10.14g
Mass of crucible lid and hydrate: 11.20g
Mass of hydrate: 1.06g
Mass of crucible, lid and an hydrous salt(1st): 10.83g
Mass of crucible, lid and an hydrous salt(2nd): 10.81g
Mass of anhydrous salt: 0.67g
Mass of water given off: 0.39g
Mass of one mole of anhydrous salt from instructor: 159.6g/mol

1. 0.39g/1.06*100%= 36.8
2. 0.67g/159.6g/mol= 0.0042mol
3.0.39g/18g/mol=0.022mol
4. moles of water/ moles of CuSO4= 0.022mol/0.0042mol= 5
5.CuSO4 5H2O

Molar Volume at STP

Molar Volume of a Gas at STP

As you may know, gases change volume (expand and contract) with changes in pressure and temperature.

To compare volume of gases we use a standard condition called STP(Standard Temperature and Pressure)

STP = 1 atmosphere of pressure and a temperature of 0C (or) 273.15K

If at STP, 1 mole of gas occupies 22.4L, therefore we can create these conversion factors:
22.4L of gas
----------------
1 mole of gas

(or)

1 mole of gas
-----------------
22.4L of gas

Examples:
1. What volume will 3.20 moles of 02 gas occupy at STP?

3.20moles x 22.4L
1 mole

3.20moles x 22.4L “moles cancel, left with litres”
1 mole

= 71.7L occupied by 3.20moles 02 gas

2. What volume will 0.500 moles of NH3 gas occupy at STP?

0.500moles x 22.4L
1 mole

0.500moles x 22.4L “moles cancel, left with litres”
1 mole

= 11.2L occupied by 0.0500moles NH3 gas

3. What volume will 60.00g of CO2 gas occupy at STP?

28.0g x 1mole x 22.4L
28g 1mol


44.0g x 1mole x 22.4L = 30.55L occupied by 60g CO2
44g 1mole
“first convert to moles, then use moles to calculate litres occupied”
= 3630L occupied by 9.758 x 1025 atoms of N2O5

Diluting Solutions

Chemicals are shipped around the world is their most concentrated forms(solids, concentrated acids, etc)


For example.You have a 5.0 mg/L phosphate standard that exceeds the upper test limit for the low-range phosphate analytical procedure. For it to be considered adequate for a QA/QC known-concentration sample you’ll need to dilute it to a 0.2 mg/L concentration. Can you easily and accurately make this dilution?

Or your 6 N sodium hydroxide solution is too strong to properly adjust the pH of BOD samples. Instead, a 1 N solution would give greater control. How much of the 6 N solution would you need to dilute to get a 1 N solution?

There are also many other instances in a wastewater laboratory where the dilution of an acid, a base, or a laboratory standard is required so that when the resulting solution is used in an analytical procedure, it will help you attain the most accurate result.

The formula for diluting these types of solutions is simple:

(volumeA)(concentrationA) = (volumeB)(concentrationB)

Here are two examples using this formula:

#1) You want to make 250 mL of a 0.20 mg/L phosphate solution from a stock solution of 5.0 mg/L. How much of the 5.0 mg/L phosphate stock solution will you need to dilute to 250 mL so that the resulting concentration is 0.20 mg/L phosphate?

Put the information into the formula: (250 mL)(0.20 mg/L) = (X mL needed)(5.0 mg/L)
Then solve for X: (X mL needed) = (250 mL)(0.20 mg/L)/(5.0 mg/L) = 10 mL

Take 10 mL of the 5.0 mg/L phosphate solution and dilute it to 250 mL with distilled water (10 mL of solution with 240 mL water). The resulting solution will be 250 mL with a concentration of 0.20 mg/L phosphate.

Molarity

Molarity is the concentration of an element or compound in water. Molarity is known as M or mol/L.
The higher the molarity the more concentrated an element or compound is.

Molarity is easily calculated by finding how many moles are present and then dividing by the volume ( in litres). Simply if there are 2.5 moles present of a compound and the volume is 10 litres the molarity is 2.5/10 or .25 M.

Empirical & Molecular and Percent Composition

Empirical fomulas are the simplified versions of compounds.
For example C3H9 can be simplified to CH3 and on the otherhand molecular formulas are the unsimplified versions of empirical compounds like C3H9

Empirical and molecular formulas can be expressed as percent compositions.
For exmaple water is 89% oxygen and 11% hydrogen. Though more hydrogen atoms are presesnt in the formula H2O the percent compisition is based on the masses of atoms divided by the total mass of the compound.

For H2O the percent composition is determined by:
Oxygen = 16 g/mol
Hydrogen = 1g/mol
Total mass = 16 + 2 (2 hydrogen atoms are present)

Oxygen % = 16/18 or 89%
Hydrogen %= 2/18 11%

With organic compounds that are burned it is possible to find out how much of each atom was present in the beginning of the reaction and after.

If a 6.5 gram sample of C and H burn to produce 20.5 grams of CO2 and 8.4 grams of H2O
You can figure out the empircal or molecular formula by finding how many moles are present of each compound.

There is .466 mol of CO2 (remember mole conversions)
And .467 mol of H2OIf you divide by the lowest mole amount you can get how many of each compound are present.
.467/4.66 = 1
.466/.466 = 1
So theres about 1 of H2O and CO2
And knowing this simple ratio of xCHy + O2 = xCO2 + (y/2)H2O
In this case it would be 1CH2 + 2O2 = 1CO2 + 1H2O
And to find the oxygen is as simple as adding the oxygens present on the the right side of the formula and dividing by two.

Mole Conversions 2 Steps

Converting can be done from grams to moles to particles and vice versa step by step or in a two step conversion.

It's possible to go from particles to grams and grams to particles using the simple mole conversions
For example:
6.24 grams of NO3  To particles.
First you would need to find the molecular mass by finding the atomic mass of each seperate element and adding them. The molecular mass in this case would be 60u.

Next you would go from grams to moles and then to particles by

6.24 g x 1 mol/60 g x 6.022x10^23/1 mol
=6.3x10^22 particles (don't forget sig figs)

Mole Conversions

Grams => Moles
= *1mol/ (?)g

Moles => Formula unit/ Particle/ Atoms
= *( 6.022*1023)/1mol

Formula unit/Particle/Atoms => Moles

=*1mol/( 6.022*1023)

Moles=> Grams
=* (?)g/ 1mol

1. How many moles of Ag are present in 3.0*1016 atoms of Ag?

3.0*1016  *1mol/( 6.022*1023)
= 5.0*10-8 Ag atoms

2.  How many atoms are present in 2 moles of carbon?

2moles C*( 6.022*1023)/1mol
= 1*1024 atoms C

3. What is mass in grams of 1.41 moles of Iron?
Atomic mass of Fe= 5.845u
Molar mass of Fe= 5.845g/mol
1.41mol Fe * 5.845g Fe/ 1mol Fe = 8.24g Fe

4. How many moles are there in 92.0 grams of lead?
Atomic mass of Pb = 207.2u
Molar mass of Pb = 207.2g/mol

92.0g Pb * 1mol Pb/ 207.2g Pb = 0.444 mol Pb