Call Now to Set Up Tutoring:

(888) 888-0446

AP Chemistry : Other Stoichiometric Calculations

Study concepts, example questions & explanations for AP Chemistry

Create An Account

All AP Chemistry Resources

6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Other Stoichiometric Calculations

You have an unidentified colorless, odorless liquid in a thin cubic container. The container is 3.2 inches by 3.4 inches by 3.3 inches and the liquid fills the entire container. The mass of the liquid minus the mass of the container is 0.370 pounds. What is the liquid's density in grams per milliliter?

(1 inch = 2.54 cm and 1 kg = 2.2046 pounds)

Possible Answers:

$0.364\frac{g}{mL}$

$0.827\frac{g}{mL}$

$0.285\frac{g}{mL}$

$1.84\hspace{1 mm}\frac{g}{mL}$

$1.32\frac{g}{mL}$

Correct answer:

$0.285\frac{g}{mL}$

Explanation:

First we must determine the volume of the container:

$3.2\hspace{1 mm}in\times 3.4\hspace{1 mm} in\times 3.3\hspace{1 mm} in= 35.9\hspace{1 mm}in^3$

The density is $\frac{0.370\hspace{1 mm}lbs}{35.9\hspace{1 mm}in^3}$ .

Now we will convert this into grams per milliliter.

$\frac{0.370lbs}{35.9in^3}*(\frac{1in}{2.54cm})^3*\frac{1cm^3}{1mL}=6.29*10^{-4}\frac{lb}{mL}$

$6.29*10^{-4}\frac{lb}{mL}*\frac{1kg}{2.2046lb}*\frac{1000g}{1kg}=0.285\frac{g}{mL}$

Report an Error

Example Question #22 : Stoichiometry

How much heat is released from the reaction of 18 moles of methane if each mole of methane liberates 5kJ of heat, but the reaction is only 60% efficient?

Possible Answers:

5kJ

90kJ

54kJ

60kJ

18kJ

Correct answer:

54kJ

Explanation:

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

$18mol * \frac{5kJ}{mol} = 90kJ$

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

$18mol * 0.60 * \frac{5kJ}{mol} = 54kJ$

Report an Error

Example Question #3 : Other Stoichiometric Calculations

When $5.884\,g$ of NaCl are dissolved in $100\,mL$ of a $1.00\,M$ solution of $AgNO_{3}$

Possible Answers:

after filtration $170\,g$ of a solid are obtained

we obtain a clear solution

the approximate concentration of $NO{_{3}}^{-}$ in the resulting solution will be $1.00\,M$ and after filtration $17.0\,g$ of a solid are obtained

a precipitate of sodium nitrate will be obtained

Correct answer:

the approximate concentration of $NO{_{3}}^{-}$ in the resulting solution will be $1.00\,M$ and after filtration $17.0\,g$ of a solid are obtained

Explanation:

The net ionic equation that occurs is:

$Ag_{(aq)}^{+}\; +\;Cl _{(aq)}^{-}\; \rightarrow \; AgCl_{(s)}$

$NO_{3}^{-}$ and $Na^{+}$ are spectator ions. Hence, $NO_{3}^{-}$ concentration is not affected. The molecular mass of NaCl is 58.44 g , then 5.844 g of NaCl yields $0.1000 \: mol$ of $Cl^{-}$ . Since we have $0.100\: L$ of a $1.00 \: M$ solution of $AgNO_{3}$ , we have $0.100\: mol$ of $Ag^{+}$ . There is no limiting reactant. Hence $0.100\: mol$ of AgCl are formed. Being the molecular mass of $AgCl \: 169.9 \: g/mol$ , $17.0 g \: of \: AgCl$ are formed. Result should be given with tree significant figures.

Report an Error

Example Question #2 : Other Stoichiometric Calculations

What is the mass of $8.888 * 10^{9} atoms \ Fe$ ?

Possible Answers:

$8.24 * 10^{-13} g\ Fe$

$2.64 * 10^{-16} g\ Fe$

$2.989 * 10^{35} g\ Fe$

$8.243 * 10^{-13} g\ Fe$

$2.99 * 10^{35} g\ Fe$

Correct answer:

$8.243 * 10^{-13} g\ Fe$

Explanation:

$8.888 * 10^{9} atoms \ Fe * \frac{1 mol Fe}{6.022 * 10^{23}atoms\ Fe} * \frac{55.85 g Fe}{1 mol Fe} = 8.243 * 10^{-13} g Fe$

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.

Report an Error

Example Question #3 : Other Stoichiometric Calculations

How many atoms are there in 17.0g of sulfur?

Possible Answers:

$8.81 * 10^{-25} atoms\ S$

$3.28 * 10^{26} atoms\ S$

$3.3 * 10^{26} atoms\ S$

$3.2 * 10^{23} atoms\ S$

$3.19 * 10^{23} atoms\ S$

Correct answer:

$3.19 * 10^{23} atoms\ S$

Explanation:

$17.0 g\ S * \frac{1 mol S}{32.06 g\ S} * \frac{6.022^{23} atoms \ S}{1 mol\ S} = 3.19 * 10^{23} atoms \ S$

Remember, there really isn't a way to go straight from grams to atoms, but, it's possible to change grams into moles (using the molar mass), and it's possible to change moles into atoms (using Avogadro's number).

Report an Error

Example Question #31 : Stoichiometry

For the molecular formula, $CH_{3}CH_{2}Br$ , how many moles of bromine are in 20.0 grams of this molecule?

Possible Answers:

1.10mol

20.55mol

0.1835mol

8.1112mol

15.0mol

Correct answer:

0.1835mol

Explanation:

The first step is to determine the molecular weight of the molecule:

$MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ Br)$

$MW_{CH_{3}CH_{2}Br}=2(12\frac{g}{mole})+5(1\frac{g}{mole})+(80\frac{g}{mole})=109 \frac{g}{mole}$

The molecular weight serves as a conversion factor to convert grams to moles as implied its units $\frac{g}{mole}$ . So, given that we are dealing with 20.0 grams, the moles of 20.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$20.0 g * \frac{mole}{109\ grams} =0.1835\ moles\ CH_{3}CH_{2}Br$

Based on the molecular formula, for every mole of the compound $CH_{3}CH_{2}Br$ , there is 1 mole of bromine. Based on our calculations 20.0 grams of $CH_{3}CH_{2}Br$ is 0.1835 moles of the compound. Because there is a 1:1 molar ratio of bromine to every molecule, we have 0.1835 moles of bromine in 20 grams of our compound, $CH_{3}CH_{2}Br$ .

Report an Error

Example Question #32 : Stoichiometry

For the molecular formula, $CH_{3}CH_{2}OH$ , how many grams of oxygen are in 50.0g of this molecule?

Possible Answers:

17.4g

16.2g

46.1g

12.4g

20.7g

Correct answer:

17.4g

Explanation:

The first step is to determine the molecular weight of the molecule ethanol, $CH_{3}CH_{2}OH$ :

$MW=\text{respective equivalents *[(atomic weight of C)+(atomic weight of H)+(atomic weight of O)]}$

$MW_{CH_{3}CH_{2}OH}=2\left(\frac{12g}{mol}\right)+6\left(\frac{1g}{mol}\right)+\left(\frac{16g}{mol}\right)=46 \frac{g}{mol}$

The molecular weight serves as a conversion factor to convert grams to moles as implied by its units $\frac{grams}{mole}$ . So, given that we are dealing with 50 grams of ethanol, the moles of 50 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$50g * \frac{1mole}{46 g} =1.09 mol\ CH_{3}CH_{2}OH$

Based on the molecular formula, we can see that for every mole of the compound $CH_{3}CH_{2}OH$ , there is 1 mole of oxygen. Based on our calculations, 50 grams of $CH_{3}CH_{2}OH$ is 1.09 moles of the compound. Because there is a 1:1 molar ratio of oxygen to every ethanol molecule, that is, 1 mole of oxygen in every mole of our molecule, we have 1.09 moles of oxygen in 50 grams of our compound $CH_{3}CH_{2}OH$ . To convert 1.09 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which from the periodic table we know is $16\frac{g}{mol}$ :

$1.09 mol * \frac{16g}{mol}= 17.4 g\ oxygen$

Report an Error

Example Question #6 : Other Stoichiometric Calculations

For the molecular formula, $C_{4}H_{9}O_{2}$ , how many grams of oxygen are in 45.0 grams of this molecule?

Possible Answers:

10.1g

25.4g

11.4g

55.0g

16.2g

Correct answer:

16.2g

Explanation:

The first step is to determine the molecular weight of the molecule:

$MW=\text{respective equivalents *[(atomic weight of C)+(atomic weight of H)+(atomic weight of O)]}$

$MW_{C_{4}H_{9}O_{2}}=4(12\frac{g}{mole})+9(1\frac{g}{mole})+2(16\frac{g}{mole})= 89 \frac{g}{mole}$

The molecular weight serves as a conversion factor to convert grams to moles as implied its units $\frac{g}{mole}$ . So, given that we are dealing with 45.0 grams, the moles of 45.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$45.0g * \frac{mole}{89.0\ grams} =0.506\ moles C_{4}H_{9}O_{2}$

Based on the molecular formula, we can see that for every mole of the compound $C_{4}H_{9}O_{2}$ , there are 2 moles of oxygen. Based on our calculations 45.0 grams of $C_{4}H_{9}O_{2}$ is 0.506 moles of the compound. Because there is a 2:1 mole ratio of oxygen to every molecule, that is, 2 mole of oxygen in every mole of our molecule, we have double the moles of oxygen in our compound $C_{4}H_{9}O_{2}$ as calculated below:

$0.506\ moles_{C_{4}H_{9}O_{2}} * \frac{2\ moles\ of\ oxygen}{1\ mole_{C_{4}H_{9}O_{2}}}=1.011\ moles\ of\ oxygen$

To convert 1.011 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which is $16\frac{g}{mole}$ :

$1.012\ moles * \frac{16g}{mole}= 16.2 g\ oxygen$

Report an Error

View AP Chemistry Tutors

Ekta
Certified Tutor

Punjab Agricultural University, India, Doctorate (PhD), Chemistry.

View AP Chemistry Tutors

Daniel
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Psychology.

View AP Chemistry Tutors

Alejandro
Certified Tutor

Kennesaw State University, Bachelor of Science, Computer Software Engineering.

All AP Chemistry Resources

6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Philadelphia, GRE Tutors in Houston, French Tutors in Atlanta, Chemistry Tutors in Atlanta, LSAT Tutors in Miami, MCAT Tutors in Phoenix, Physics Tutors in Miami, Biology Tutors in Dallas Fort Worth, Biology Tutors in Seattle, Biology Tutors in Phoenix

Popular Courses & Classes

GMAT Courses & Classes in New York City, ISEE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Chicago, LSAT Courses & Classes in Philadelphia, SSAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Washington DC, ACT Courses & Classes in Houston, LSAT Courses & Classes in Seattle, ISEE Courses & Classes in Houston

Popular Test Prep

GMAT Test Prep in Los Angeles, ISEE Test Prep in Boston, SSAT Test Prep in Washington DC, SSAT Test Prep in New York City, ISEE Test Prep in Seattle, ISEE Test Prep in Houston, MCAT Test Prep in Philadelphia, MCAT Test Prep in Los Angeles, GRE Test Prep in Dallas Fort Worth, GRE Test Prep in Washington DC

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Chemistry : Other Stoichiometric Calculations

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Other Stoichiometric Calculations

Example Question #22 : Stoichiometry

Example Question #3 : Other Stoichiometric Calculations

Example Question #2 : Other Stoichiometric Calculations

Example Question #3 : Other Stoichiometric Calculations

Example Question #31 : Stoichiometry

Example Question #32 : Stoichiometry

Example Question #6 : Other Stoichiometric Calculations

All AP Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: