Moles Molecules And Grams Lab 3b3gy

Chemistry

Name: Henry Pontak Period: 1 Date: November 28, 2013

Moles, Molecules, and Grams Lab HP T h e

M o l e

C o n c e p t

U n i t

4

As we’ve already discussed in class, it’s easy to make conversions between moles, molecules/formula units, and grams. For example, if we want to go from moles to grams, we use the molar mass to make this conversion. If we want to go from moles to molecules, we use Avogadro’s number, or 6.02 x 1023. Finally, if we want to go from grams to molecules/formula units, we just use a two step process where we first convert from grams to moles, and then from moles to molecules/F.U. Visually, it looks like this: Molar mass Grams

Moles

6.02 x 1023

Molecules/Formula Units

In this lab, we will be weighing out five different substances then finding out how many moles and molecules/F.U. of each one we have. Prelab: If you measure out 25.0 grams of NaOH in this lab, how many moles of NaOH would you have? How many formula units? 25.0 g NaOH x 1mol/39.997g = 0.63 mol of NaOH

Lab: In this lab, there are five different balances set up, each of which has a labeled substance in a little jar next to it. Your job is to find out how many moles and how many molecules of each substance are in the canister. Some     

information you might find useful: Sand (SiO2) baking soda (NaHCO3) chalk (CaCO3) table salt (NaCl) sugar: sucrose (C12H22O11)

Make    

sure that your lab has a data table that contains the following: The name of each of these substances The formula for each of these substances The molar mass of each of these substances The mass in grams of each of these substances.

When you are done weighing each material, calculate the number of moles of each material, and the number of molecules of each one, and put them in the right space. All calculations must be shown on lined paper with all conversion factors and unit cancelling.

Data Table: Formula & Name

Sand (SiO2) Silicon Dioxide

Baking Soda (NaHCO3) Sodium Bicarbonate

Salt (NaCl) Sodium Chloride

Sugar (C12H22O11) Sucrose

84.007g

Chalk (CaCO3) Calcium Carbonate 100.0869g

Molar Mass in g Mass of sample in g Formula for moles Number of Moles Formula for molecules

60.08g

58.44g

342.2965g

2.961g

2.546g

0.75g

3.7g

2.013

2.961g x 1mol/60.08g 0.04927 mol

2.546g x 1mol/84.007g 0.03031 mol

0.75g x 3.7g x 2.103g x 1mol/100.0869g 1mol/58.44g 1mol/342.2965g 0.0075 mol 0.063 mol 0.005881 mol

0.04927mol x 1mol/ 6.02x1023 molecules

0.03031mol x 1mol/ 6.02x1023 molecules

0.0075mol x 1mol/ 6.02x1023 molecules

Number of Molecules

8.184x1026 molecules

5.035x1026 molecules

1.2x1026 molecules

0.063mol x 1mol/ 6.02x1023 molecules 1.0x1027 molecules

0.005881mol x 1mol/ 6.02x1023 molecules 9.769x1027 molecules

Calculations: Using your data, find the following values. Make sure to show all work and write all numbers with the correct significant figures. Work is above in data table!! 1a. Number of moles of sand: 0.04927 moles

b. Number of molecules of sand: 8.184x1026 molecules

2a. Number of moles of baking soda: 0.03031 moles

b. Number of formula units of baking soda: 5.035x1026 Formula Units

3. Number of formula units of chalk: 1.2x1026 Formula Units Hint: String two conversions together

4. Number of formula units of salt: 1.0x1027 Formula Units

5. Number of molecules of sugar: 9.769x1027 molecules

Post Lab Questions: 1. Which of the materials we worked with had the largest number of molecules/formula units? Was this the material that had the largest weight? Why? Sugar had the largest number of molecules with a number of 9.769x1027 molecules. This was not the sample that had the largest weight however; it did have a really high molar mass with a number of 342.2965g. Since sugar had a really high molar mass, this element must have a high number of moles. Therefore having a really high number of molecules.

2. Water has a molecular formula of H2O. If I have 50.0 g of water, how many moles of water do I have? How many molecules? Molar Mass of H20= 18.01528 50.0 g H20 x 1mol/ 18.01528g = 2.78 moles of H20 2.78mol H20 x 6.02x1023/1 mol = 1.67x1024 molecules of H20

3. Butane has a molecular formula of C4H10. If I have 50.0 grams of butane, how many moles of natural gas do I have? How many molecules? Molar Mass of C4H10= 58.12g

50.0 g C4H10 x 1mol/58.12g = 0.86 moles of C4H10 0.86mol C4H10 x 6.02x1023/1mol = 5.18x1024 molecules of C4H10

4. I had the same weight of water and butane in problems 3 and 4. Why didn’t the answer come out the same? Explain. Even though you have the same weight of two elements, their molecules and moles will not be the same because H20 and C4H10 have different molar masses. Thus when first trying to find the moles of the sample, one must input the molar mass into dimensional analysis thus the answers will not come out the same.