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Example Questions
Example Question #1 : Molecular Weight, Molecular Formula, And Moles
The atomic mass of lithium is . What is the percent composition of lithium by isotope, assuming that its only isotopes are and ?
The atomic mass of an element is determined by the proportional mass of each elemental isotope. We know that there are only two isotopes of lithium; therefore, their percentages must add to 100%.
The atomic mass will be equal to the mass of each isotope multiplied by its abundance.
We can substitute an algebraic expression to solve for one of our variables.
Using this value, we can solve for the abundance of the other isotope.
Converting these values to percentages gives us our final answer.
Example Question #1 : Molecular Weight, Molecular Formula, And Moles
How many carbon atoms exist in two moles of carbon dioxide?
1.2 * 1024 atoms
6.022 * 1023 atoms
2.7 * 1022 atoms
32 atoms
1.2 * 1024 atoms
It is important to remember that one mole of an element contains 6.022 * 1023 atoms. Since there are two moles of carbon dioxide (CO2), we can conclude that there are two moles of carbon. As a result, there are atoms.
Example Question #1 : Stoichiometry And Analytical Chemistry
A researcher performs an elemental analysis on a compound. He finds that the compound is made up of only carbon, hydrogen, and oxygen atoms. He isolates a pure sample of the compound and finds that this sample contains of carbon, of hydrogen, and of oxygen. The researcher wants to perform further analysis on this compound the next day. Before leaving the lab the researcher creates three stock solutions of varying concentrations of this compound: (solution A), (solution B), and (solution C). He stores these solutions overnight at a temperature of .
Molecular weight of this compound =
The researcher obtains a sample of of carbon-12. How many atoms of carbon-12 are present in this sample?
A mole is defined as the number of atoms present in of carbon-12. Obtaining a sample of carbon-12 will give you of carbon-12. You can also find the number of moles by multiplying the mass by the molecular weight of carbon:
Remember that a mole of any element contains atoms (Avogadro’s number). Since we have exactly one mole in the sample, there will be exactly atoms of carbon-12.
Example Question #4 : Molecular Weight, Molecular Formula, And Moles
A researcher performs an elemental analysis on a compound. He finds that the compound is made up of only carbon, hydrogen, and oxygen atoms. He isolates a pure sample of the compound and finds that this sample contains of carbon, of hydrogen, and of oxygen. The researcher wants to perform further analysis on this compound the next day. Before leaving the lab the researcher creates three stock solutions of varying concentrations of this compound: (solution A), (solution B), and (solution C). He stores these solutions overnight at a temperature of .
Molecular weight of this compound =
When calculating the empirical formula, if you used ratios of the number of atoms of each element instead of ratios of moles of each element, would you get a different answer?
Yes, because Avogadro’s number is different for each element
Yes, because there is a constant relationship between mass and atoms of each element
No, because the number of moles and the number of atoms in each element is equal
No, because there is a constant relationship between moles and atoms of each element
No, because there is a constant relationship between moles and atoms of each element
Remember that the empirical formula relies on the ratio of the moles of elements. To get the number of atoms, you would have to multiply the moles of each element by the Avogadro’s number (). You would use this number for every element (Avogadro’s number doesn’t change for each element). This means that the ratio of the number of atoms will be the same as the ratio of the number of moles, and you will get the same empirical formula. There is a constant relationship between the number of moles and the number of atoms in a sample.
Example Question #1 : Molecular Weight, Molecular Formula, And Moles
Which of the following contains the greatest number of moles?
Each answer choice is presented in grams. To convert to moles, we will need to divide each choice by the atomic mass of the presented element.
Two grams of hydrogen atoms will result in the greatest amount of moles.
Example Question #2 : Molecular Weight, Molecular Formula, And Moles
Which of the following cannot be directly related to Avogadro's number by stoichiometry?
Partial pressure of a sample
Moles in a sample
Grams in a sample
Number of atoms in a diatomic gas sample
Number of molecules in a sample
Partial pressure of a sample
Avogadro's number is used to convert between moles and atoms (or molecules).
This immediately eliminates two answer choices, since moles and molecules in a sample are both contained in the constant. Atoms in a sample of diatomic gas is also easily related to Avogadro's number, as each molecule in the sample will contain exactly two atoms. Avogadro's number can be used to determine the number of moles, molecules, or atoms in a sample of diatomic gas.
Converting grams in a sample to moles allows us to use Avogadro's number to further convert to molecules.
Partial pressure of a gas is directly related to mole fraction, but cannot be used to determine the moles of gas unless the total moles in the sample is known. Partial pressure represents a relationship between the sample and its particular environment, whereas constants govern conversion between moles, grams, atoms, and molecules. Since partial pressure is not directly related to these terms by a constant, Avogadro's number cannot be applied to partial pressure to determine any useful data.
Example Question #1 : Molecular Weight, Molecular Formula, And Moles
Consider the following molecular formulas:
*The IUPAC name for DEET is N,N-diethyl-meta-toluamide
DEET has a density of . How many nitrogen atoms are found in one liter of DEET?
To solve, we will need to find the mass of one liter of DEET and use the mass percentage of nitrogen to find the mass of nitrogen represented. Then, the atomic mass of nitrogen can be used to convert this to moles, and Avogadro's number can be used to convert to atoms.
First, find the mass of DEET in a one-liter sample:
Find the mass of nitrogen in 998 grams of DEET by finding the percentage of nitrogen by mass:
Use this mass percentage and atomic mass to find the moles of nitrogen in the one-liter sample.
Use Avogadro's number to find the number of atoms in this sample.
An alternative method would be to convert the grams of DEET to moles, and then from moles to molecules. There is one nitrogen atom per molecule of DEET, so this would method also get you the same answer.
Example Question #2 : Molecular Weight, Molecular Formula, And Moles
Which of the following samples contains atoms?
All samples of equal mass have the same amount of atoms
The number of atoms in a sample is dependent on the molar quantity, NOT the mass of the sample. Avogadro's number shows that there are atoms in a mole of any element. Knowing this, we can find out how many moles of the element are represented by the given amount of atoms.
By dividing the mass of the samples by the molar mass of each respective element, we can find which sample has this molar quantity.
Sodium:
Lithium:
Potassium:
As a result, a five-gram sample of sodium has .
Example Question #1 : Molecular Weight, Molecular Formula, And Moles
How many sodium ions are present in of a solution of sodium hydroxide?
A full liter of a one molar solution of sodium hydroxide would contain one mole of sodium ions, or ions. Here, you have only one tenth the volume, so multiply the number in one mole by one tenth.
Now that we have reduced the volume, we need to account for the concentration.
Example Question #10 : Molecular Weight, Molecular Formula, And Moles
What is the empirical formula for a compound that contains 41.4% carbon, 6.9% hydrogen, 27.6% oxygen, and 24.1% nitrogen?
C3H5ON2
C2H6O2N2
C2H4ON
CH4O3N2
C2H4ON
When finding the empirical formula for a compound, it helps to imagine a 100g sample of the molecule. This way, the percentages of the atoms can be converted to amounts in grams. At this point, dividing each amount by the atom's molar mass results in the values that will be compared to one another in order to find the ratio of atoms in the molecule.
(carbon)
(hydrogen)
(oxygen)
(nitrogen)
From these calculations, we can set up a ratio.
3.45C : 6.9H : 1.72O : 1.72N
If we divide this ratio by the smallest value (1.72), we can see that it can be reduced.
2C : 4H : 1O : 1N
As a result, the empirical formula for the compound is C2H4ON.
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