Question

Option 2:  Kim thinks she’s got Avogadro’s number down, she can use it to get to molecules from moles and to moles from molecules, but she runs into a roadblock when they start using atoms within the molecules.  She comes to you with the following problems and wants a step by step explanation of how to solve them.  What conversions would you have her use?  What steps are involved for each one?  How would you explain how these are different from the problems she already understands?

If you have 7.54 x 10²⁵Oxygen atoms in Silver sulfate, what mass of silver sulfate do you have?

If you have 2.89g of Aluminum Hydroxide.  How many moles of Hydrogen do you have in the Aluminum Hydroxide?  How many atoms of Hydrogen is this?

If you have 70.0g of Chloride ions in Calcium chloride, how many molecules of calcium chloride do you have?

Option 3:  Jake is confused about where to start in solving questions about moles, grams, and molecules. He brings you the following problems. Can you teach him HOW to tell these four things for problems A-C?

What the starting information is in each problem.

What they question wants him to solve for.

When to use which conversions.

How to set up the math for each problem

Then walk him through to the solution of each problem.  The problems he is looking to solve are:

How many moles are found in 65.7g of Lithium Sulfate?

How many moles of H are found in 22.5g of Ammonium Acetate?

What mass of water do you have if you have 6.43 x 10²⁸molecules of water?

Answer 1

1)

Silver sulfate is Ag2SO4 so, number of moles of Ag2SO4 = no. of moles of O atoms/4 (Each molecule contains 4 O atoms)

So, 7.54*10²⁵ O atoms have 7.54*10²⁵/Na moles of atoms = 7.54*10²⁵/6.022*10²³ = 125.2 moles of O atoms so, moles of Ag2SO4 = 125.2moles/4 = 31.3 moles

Molar mass of Ag2SO4 = molar mass of Ag*2 + molar mass of S + molar mass of O*4 = 107.8g/mole*2 + 32g/mole + 16g/mole*4 = 311.6 g/mole

So, weight of 31.3 moles of Ag2SO4 = moles* molar mass = 31.3moles*311.6g/mole = 9753 g

2)

Aluminium hydroxide is Al(OH)3 and its molar mass = molar mass of Al + molar mass of O*3 + molar mass of H*3 = 27g/mole + 16g/mole*3 + 1g/mole*3 = 78 g/mole

So, moles of Al(OH)3 in 2.89 g = weight/molar mass = 2.89g/ 78g/mole = 0.037 moles

As each Al(OH)3 molecule contains 3 atoms of H we can say that each mole of Al(OH)3 will contain 3 moles of H atoms so, 0.037 moles of Al(OH)3 will contain 0.037*3 moles of H atoms = 0.111 moles

3)

Atoms in 0.111 moles = moles*Avogadro's constant = 0.111*6.022*10²³ = 6.69*10²³ H atoms.

4)

Chloride ions are Cl- ions with molar mass = 35.45 g/mole

So, moles of Cl- in 70 g = weight/molar mass = 70g/ 35.45g/mole = 1.97 moles

Calcium chloride is CaCl2

Each molecule of CaCl2 contains two chloride ions (2Cl-) so, 1.97 moles of Cl- will be present in 1.97/2 moles of CaCl2 = 0.987 moles

Molecules of CaCl2 in 0.987 moles = moles*Avogadro's constant = 0.987*6.022*10²³ = 5.94*10²³ molecules