# CHEM125 – CHAPTER 03 – BASIC HOMEWORK I

## CHAPTER 3 BASIC HOMEWORK I

### Resources:

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### Question 1

What is the mass of 1 mol of CH3NH2 in g?

Help: Molar mass ### Explanation

1 mol = x g / MM CH3NH2 MM of CH3NH2 = 12 + 5(1.008) + 14 = 31 1mol= x g / 31 MM x= 31 g

### Question 2

What is the percent composition by mass of oxygen in carbon dioxide?

Help: Percent composition by mass ### Explanation

carbon dioxide = CO2 C= 12 MM O =16 MM * 2 = 32 32/ (32+12) = .7272 * 100 = 72.72%

### Question 3

What is the percent composition by mass of carbon in lithium carbonate, Li2CO3? ### Explanation

Li2CO3 Li-6.941(2)=13.882 C-12 O-16(3)=48 12/ (13.882 +12 +48) = 16.22

### Question 4

What is the mass of 41 molecules of nitrogen dioxide, NO2, in g? USE 3 SIG. FIG.

Help: Conversion Factor ### Explanation

(14 +16(2)) MM NO2 41/(6.022*10^23) = 6.81e-23 mol = X g/ 46 MM X= 3.1326e-21

### Question 5

What is the mass of 3.66 mol NaCl in grams? Help: Moles to mass ### Explanation

3.66 mol = X g/ MM of NaCl MM of NaCl = 58.4428 3.66 mol = X g/ 58.4428 MM X = 213.9

### Question 6

What is the mass in grams of 0.13 mol sodium phosphate, Na3PO4 ? ### Explanation

0 .13 mol = X g/ MM of Na3PO4 MM of Na3PO4 = 163.94 0.13 mol = X g/ 163.94 MM X = 21.3122

### Question 7

How many moles is 390.0 g calcium carbonate, CaCO3 ? Help: Grams to moles ### Explanation

390 g/ MM of CaCO3 = X moles MM of CaCO3 = 100.0869 390/ 100.0869 = 3.9 X = 3.9 moles

### Question 8

How many moles is 126 g lithium carbonate, Li2CO3 ? ### Explanation

126 g Li2CO3/ 73.891 MM 1.7 moles Li2CO3

### Question 9

What is the empirical formula for acetone, C2H4O2? (Keep elements in same order for your answer) Remember to use correct symbols for elements–C for carbon (NOT c), etc. Help: Empirical formula explanation

### Explanation

The empirical formula indicates the simplest whole number ratio of atoms in a substance. The molecular formula indicates the number of atoms of each element in a molecule of the substance. Ex: Molecular formula: H2O2; empirical formula: HO Molecular formulas: C2H4, C5H10; empirical formula: CH2 Molecular formula: CH4; empirical formula: CH4 C2H4O2 / 2 CH2O

### Question 10

A substance contains 5.00 g C and 1.70 g H. What is the empirical formula? (Answer starts with C and H is last.) Help: Empirical formula from mass

### Explanation

5 g C 5 g / 12 MM = .4167 mol C 1.70 g H / 1.008 MM = 1.686 mol H 1.686 mol H / .4167 = 4.05 about 4 CH4

### Question 11

What is the empirical formula of a compound which is 82.64 % C by mass and 17.36 % H. (Answer starts with C and H is last.) Help: Empirical formula from percent

### Explanation

Use a 100g sample… 82.64 g C / 12 MM = 6.89 17.63 g H/ 1.008 MM = 17.49 17.49 / 6.89 = 2.5 (not a whole number) multiply it by 2 to make it a whole number CH2.5 goes to C2H5

### Question 12

Balance the following chemical reaction using smallest whole numbers: C3H8(g) + O2(g) –> CO2(g)+ H2O(g) What is the number of CO2 molecules in the balanced equation? ### Explanation

WHOOT BALANCING Left side v Right Side Left

• 3 mols C
• 8 mols H
• 2 mols O

Right

• 1 mol C
• 3 mol O
• 2 mol H

Lets start by balancing the H… simply add a 4 coefficient in front of the H2O on the right C3H8(g) + O2(g) –> CO2(g)+ 4H2O(g) Next the Cs… add a 3 coefficient in front of the CO2 on the right C3H8(g) + O2(g) –> 3CO2(g)+ 4H2O(g) Next the Os… add a 5 coefficient in front of the O2 on the left C3H8(g) + 5O2(g) –> 3CO2(g)+ 4H2O(g) Now lets recheck to make sure 3 C: L and R 8 H: L and R 10 O: L and R

### Question 13

What is the number of H2O molecules in the balanced equation? Look Above

### Question 14

What is the number of O2 molecules in the balanced equation? ### Question 15

Balance the following chemical reaction using smallest whole numbers: CuO(s) +NH3(g) –> Cu(s) + N2(g) + H2O(g) Hint, balance the elements in the following order–nitrogen, hydrogen, oxygen, and copper. What is the number of Cu atoms in the balanced equation? ### Explanation

Balancing equations heck friggin yeah… Basically all you need to do is get the same amount of stuff on both sides On the reactants side aka the left side we have: 1 mol of Cu 1 mol of O 1 mol of N 3 mols of H On the products side aka the right side we have: 1 mol of Cu 1 mol of O 2 mols of N 2 mols of H … so were going to balance out the N first and put a coefficient of 2 in front NH3 (on the reactant side) so we can have 2 mols of N on both sides

• when we do this we double the mols of H on the left

…next we balance the H and were going to put a 3 in front of the H2O on the right side so we can have 6 mols of H on both sides– next we balance the Owe have three mols on the right so were going to put a 3 coefficient on the left side in front of the CuO to balance the Os..and finally we balance the Cuand all we have to do is put a 3 coefficient in front of the Cu on the right side and there we have it

### Question 16

What is the number of nitrogen molecules in the balanced equation? (Refer to the previous question) Nitrogen molecules refer to N2 (btw)

### Question 17

What is the number of water molecules in the balanced equation? (Refer to previous questions) Look at solutions of questions 15