# Homework 7

- Page ID
- 28864

5.107, 2.23

Question 2.23:

1. The density of liquid gold (Au) is 19.3g/cm^{3}.

a. If given a volume of 450 mL of this liquid, what is the mass?

b. If given a mass of 24.4 kg of this liquid, what is its volume in L?

Solutions For Part a:

First we must start with how to approach the problem.

1. We are given the density. So we must recall the formula we learned this year for density.

- Density= Mass(g)/ Volume(mL)

2. Part a of the question asks for us to solve for the mass. Therefore, we must rearrange the previous equation and solve for mass. We do this by multiplying both side by volume. This leaves you with the following equation:

- Mass(g)= Density(g/mL) x Volume(mL)

3. Before plugging in all of our data into this equation we must make sure that all of our units match up in order to be able to perform demensional analysis. We see that the density is given to us in g/cm^{3 }however, we are given a volume of liquid in mL. Therefore we must convert cm^{3} to mL. We must recall that 1mL=1cm^{3}. This makes our conversion very simple:

-1g/cm^{3}= 1g/mL

-19.3g/cm^{3}= 19.3g/mL

4. Now that all of our units are in alignment we can go ahead and plug all of our data into the equation solved for mass and solve the problem. Our density is 19.3 g/mL and our volume is 450 mL. When we multiply these two numbers we get an answer of 8685. Our units also end up cancelling out because we have 1/mL x mL. This leaves us with grams on the top of the equation. Our answer is then 8685 g. We are not finished yet because we have to take into account significant figures. The rule for multiplication is that there are as many significant figures in the answer as there are in the value in the equation with the least number of significant figures. In this problem, this is 450, having two significant figures. Therefore, our final answer will also have two significant figures. Since our number is 8685, we round up to 8700 giving us only two significant figures and our final answer of 8700g.

M=D x V

M= 19.3g/mL x 450 mL

M = 8685 g

M= 8700g

Solutions For part b:

1. We are given the mass and density and are asked to solve for the volume. So we must again recall the formula of density and rearrange it, in order that it will solve for volume. We do this by multiplying both sides of the equation by volume. Once this is done, in order to isolate volume on one side of the equation we divide both sides of the equation by density to leave us with an equation solved for volume.

D(g/mL) =M(g) / V(mL)

D x V= M

V = M / D

2. Now that we have an equation that solves for what we want, we now need to look at our units. We are given the mass in kilograms and the density in g/cm^{3}. First we must convert kilograms to grams in order that when we divide. The two will cancel each other out when we divide and this will leave us with volume in the units of milliliters. Then we must convert g/cm^{3} to g/mL.

Hints:

1 cm^{3}= 1mL

1000 g = 1kg

Solution:

24.4kg x 1000g/ 1kg = 24400g

1cm^{3}= 1mL

19.3 g/cm^{3}= 19.3 g/mL

3. We now have all of our units in line and we are able to plug in the values into the equation and solve for volume.

V= M / D

V= 24400 g / 19.3 g/mL

V= 1264.2487mL

4. In order to complete this problem we have two last steps. The first thing we must take into account after doing division is significant figures. The rule for significant figures when dividing is that the answer will have as many significant figures as the number in the problem with the least amount of significant figures. So we must find the number with the least amount of sig figs and then apply this number to our final answer

Hints:

trailing zeros do not count as significant figures

-5 and above rounds up, 4 and below rounds down

Solution:

24400/19.3=1264.2487

=1260 mL

5. Our last step in the problem is to make sure our answer is what the question is asking for. when we go back and look at the problem, it asks for us to solve for the volume in Liters. By looking at our dimensional analysis, we end up with volume in milliliters. We must convert this to Liters an then our answer will be complete.

Hint:

1000mL= 1L

Solution:

1260 mL x 1L / 1000 mL= 1.26 Liters

Question 5.107

After the combustion of a hydrocarbon, we find 66.02g of CO_{2} and 27.02g of H_{2}O. What is the empirical formula of this hydrocarbon?

Solutions to Question:

66.02 x 1 mol/ 44.0 g= 1.500 mol CO2= 1.5 mol C

27.02 x 1 mol/ 18g = 1.501 mol H20= 3 mol H

C1.5 H3= C3 H6

You start by taking both, divide by the molar mass, use the mol ratio of the compound and multiple to get rid of decimals.