# 2.E: Atoms and Elements (Exercises)

- Page ID
- 78962

These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

## Q2.15

The length of a Nickel coin is measured using a ruler with markings at every 0.1 mm. What is the correct measurement report for the Nickel?

**What We Know:** According to the US government, the size of a Nickel is 0.835 inches or 21.21 mm

**What It's Asking For:** Using significant figures, determine the correct measurement of the Nickel.

**Strategy:** Every digit in a scientific measurement is certain except for the last digit. Unless stated otherwise, the last digit is estimated with an uncertainty of +/- 1.

**Solution: **

**21.21 mm**- 21.2 mm
- 21 mm

## Q2.16

When measuring the mass of a cat that weighs around 4530 g that is placed on a scale that has markings at every 0.01 kg, which is the correct measurement reported on this scale?

- 4.5 kg
- 4.53 kg
- 5 kg
- 4.534 kg

**Strategy:**

First you have to convert grams to kg because the scale is measured in kg. Once it is concerted to kg, you have to remember that when reading a scale, you need to read on sig fig past where the markings read.

\[ 4534\;\cancel{g} \left( \dfrac{1\;kg}{1000\;\cancel{g}} \right) = 4.53\;kg\]

The grams will cross out to leave you with kg as the units. Since the the scale reads to the 0.01 kg, it would read as a clear marking at 4.53, and you would estimate the next number at the 0.001 kg. That being said the only answer that reads to that is D) 4.534 kg.

## Q2.20

A silver picture frame displaces 0.454 L of water and has a mass of 2.78 kg. Calculate the density of the silver in g/cm^{3}.

**Strategy**

- Convert L to cm
^{3}using dimensional analysis and conversion factors.\[0.454\;L \left( \dfrac{1,000\; mL }{1\; L} \right)\left( \dfrac{1 cm^{3}}{1 \;mL} \right) = 454cm^{^{3}}\] - Convert kg to g using dimensional analysis and conversion factors.\[2.78\;kg \left( \dfrac{1,000 \;g }{1\; kg} \right)= 2,780\;g\]
- Calculate the density \(\rho\).

\[\rho=\dfrac{m}{V}\]

\[\rho=\dfrac{2,780g}{454cm^{^{3}}}\]

\[\rho=6.12\dfrac{g}{cm^{^{3}}}\]

## Q2.23

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

- If given a volume of 450 mL of this liquid, what is the mass?
- 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 8700 g.

M=D x V

M= 19.3g/mL x 450 mL

M = 8685 g

M= 8700 g

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 *

## Q2.24

- Sodium Hypochlorite (bleach) has a density of 1.11 g/cm
^{3}and a volume of 48.21 mL, what is the mass? - What is the volume of 8.31 g of bleach?

Use the correct units in your answers.

**Answers**

- m= 53.51g
- V= 7.49mL

**Strategy for part A**

1. Recognize the equation for density is density (d) equals mass (m) divided by volume (V)

\[\rho =\dfrac{m}{V}\]

2. Plug in the provided information from the question above

\[1.11 \;g/cm^3 =\dfrac{m}{48.21 \;ml}\]

3. Convert cm^{3} into mL using the conversion factor 1 cm^{3}=1 mL, these units should then cancel each other out leaving only grams as the unit.

4.Now solve for \(m\).

\[m=53.51\;g\]

**Strategy for part B**

1. Use the density formula from part a to plug in the information given in the question above

\[\rho =\dfrac{m}{V}\]

2. Plug in the values from the question above

\[1.11\; g/cm^3 =\dfrac{8.31\;g}{V}\]

3. To get volume in mL, convert cm^{3} to mL use the conversion factor 1 cm^{3}= 1 mL.

4. Solve for \(V\)

\[V= 7.49\; mL\]

## Q2.25

A ship has a fuel tank that can hold 900 L of gasoline. The fuel has a density of 0.0432 g/cm^{3}. How much weight is the ship carrying in fuel?

**Strategy**

First, you want to look at what the problem has given you. Look at the units and determine if you have been given the density, the mass, or the volume. Also, check to see if any of the units need to be converted to match the units of the density formula. Then you are going to plug the numbers you have into the formula and solve for what is being asked for.

**Solving**

The problem has given us the volume and the density of the gasoline, so now we must solve for the mass of the fuel. First, we must make sure that none of the numbers need to be converted to different units, but they do! The 900 L needs to be converted to cm^{3}. First we must convert L to mL.

\[900 \; \cancel{L} = \dfrac{1000\; ml}{1\;\cancel{L}}\]

which gives us 900,000 mL. Now mL=cm^{3} so you do not need to convert for that. So now we have 900000 cm^{3} and we can plug all our numbers into the density formula. The density formula is

\[\rho=\dfrac{mass}{volume}\]

However, since we are looking for the mass we will switch the formula around and use it in this manner

\[Mass=Density*Volume\]

So now we will plug our values into the second formula

\[Mass=\left ( 0.432 g/cm^3 \right )*\left ( 900000 cm^{3} \right )\]

After the cm^{3} cancel each other out, we are left with \(3.8 \times 10^4\) grams of gasoline.

## Q2.40

A children's cough syrup contains 25.0mg of Benzocaine (a numbing agent) per 5.00 milliliters of syrup. The bottle instructs a parent to give a max of 5.00mg per kilogram of body weight per day. How many mL of the cough syrup can a 32.0 lb toddler take in one day?

**Strategy**

**What do we know?**

- The cough syrup is concentrated at 25.0mg Benzocaine per 5.0 mL► 25.0mg/5mL. We can divide by 5 to simplify this ratio so we know how many mL contain 5.0mg, the recommended per kg dose. ► 5.0mg/1mL
- The recommended maximum dose is 5.0mg per kilogram of the child's weight. ► 5.0mg/kg

**What are we trying to find?**

- The amount of mL needed to supply 5mg per kilogram of body weight.
- Child's weight in kg

Since the ratio of Benzocaine to kilogram of weight is 5:1, we can find the weight in kilograms and use dimensional analysis to find the correct ratio.

**Solving:**

**A **To convert, we can use dimensional analysis. Remember that each "fraction" represents a ratio of two things to each other. In order for dimensional analysis to work, units for the given information have to match the units in the denominator of the ratio to the right of it. (Think of it this way, we cant divide apples by oranges!)

\[32\;lb \cdot \dfrac{1lb}{2.204kg}= 14.519\;kg\]

**B **When calculating sig figs, we round our answer at the end. We are going to use this number in our dimensional analysis question.

**C **Using dimensional analysis, we can multiply the weight by the ratio of Benzocaine per kg. The kg unit cancels because it is present both as a numerator and denominator. Cancelling them is equivalent to simplifying them to an implied 1. The units on the far right should be the answer we are looking for.

\[14.519kg \cdot \dfrac{5mL}{1kg}= 72.595mL\]

**Answer**: The child can take a maximum of 72.6mL of the cough syrup.

## Q2.45

Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or compound. If it is a mixture, classify it as homogenous or heterogeneous.

- tears
- lithium hydroxide
- gold
- bowl of chili

**Strategy**

- You must know the difference between a:
- A substance that is made up of only one type of atom or molecule.*pure substance*- A material system made up of two or more different substances that are mixed but not combined together chemically.__mixture__

- You must know the difference between:
- Cannot be broken down into simpler substances.*element*- Substance formed when two or more chemical elements are bonded together chemically and can be broken down into simpler substances.__compound__- Mixture that has a uniform composition and properties throughout (cannot see ingredients)*homogeneous*- Mixture that remains physically separated throughout (can see ingredients)*heterogeneous*

- Label either pure substance or mixture
- Label each pure substance as either an element or compound and each mixture as either homogeneous or heterogeneous.

**Solution**

*homogeneous mixture**pure compound**pure element**heterogeneous mixture*

## Q2.47

How many Magnesium atoms are there in 4.67 mol of Magnesium?

**What we know:** We know that we have 4.67 mol of Magnesium.

**What we are asked for:** The number of Magnesium atoms present.

**Strategy:**

- Familiarize yourself with Avogadro's number which is 6.022 x 10
^{23}. - Set up a dimensional analysis equation to convert Magnesium moles to Magnesium atoms.
- Carry out the equation.

**Solution:**

- 1 mol is equal to 6.022 x 10
^{23}of anything, but in this case Magnesium atoms. - \[4.67 mol Mg \times \dfrac{6.022\times 10^{23} atoms Mg}{1 mol Mg}\]
- The "mol Mg" units cancel so you're left with:

\[4.67 \times (6.022\times 10^{23} atoms Mg) = 2.81\times 10^{24} atoms Mg\]

2.81 x 10^{24} atoms of Magnesium are in 4.67 mol of Magnesium

## Q2.55

Find the mass (in grams) of each of the following:

- 2.1 x 10
^{22 }Fe molecules - 2.9 x 10
^{23}Al molecules - 3.9 x 10
^{22}Ca molecules - 4.3 x 10
^{22 }Ga molecules

**Solution**

- Firstly, because the question requires the same steps for each part, the best way to begin would be to break down the steps necessary into a basic equation:

\[\dfrac{number\; of\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{g}{1\; mol}\]

This equation is formed by the fact there are 6.022•10^{23} molecules per mole. Next, the g/mol is the atomic weight of the respective elements in order to finalize the conversion from molecules to grams.

- Next is to fill in the numbers for each part and each element.

a. \[\dfrac{2.1\cdot 10^{22}Fe\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{55.845\; g}{1\; mol}\]

\[=1.9 g\]

b.\[\dfrac{2.9\cdot 10^{23}Al\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{26.982\; g}{1\; mol}\]

\[=13.0 g ( from 12.99 g)]\]

c. \[\dfrac{3.9\cdot 10^{22}Ca\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{40.078\; g}{1\; mol}\]

\[=2.6 g\]

d. \[\dfrac{4.3\cdot 10^{22}Ga\; molecules}{1}|\dfrac{1\; mol}{6.022\cdot 10^{23}molecules}|\dfrac{69.723\; g}{1\; mol}\]

\[=5.0 g ( from 4.98 g)\]