Molecular Weight, Molecular Formula, and Moles - MCAT Chemical and Physical Foundations of Biological Systems
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What is the percentage by weight of sodium in sodium sulfate?
What is the percentage by weight of sodium in sodium sulfate?
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Sodium sulfate is given by the formula:
To find the percentage by weight, we will need to divide the mass of sodium in the molecule by the total molecular mass.
Convert the ratio to a percentage.
Sodium sulfate is given by the formula:
To find the percentage by weight, we will need to divide the mass of sodium in the molecule by the total molecular mass.
Convert the ratio to a percentage.
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How many sodium ions are present in 
of a 
solution of sodium hydroxide?
How many sodium ions are present in
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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.
A full liter of a one molar solution of sodium hydroxide would contain one mole of sodium ions, or
Now that we have reduced the volume, we need to account for the concentration.
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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 lithium is
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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.
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.
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What is the empirical formula of 4-octene?
What is the empirical formula of 4-octene?
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4-octene looks like this:
To get an empirical formula, we find the ratio of each element within the compound and make it as low as possible. We have eight carbons and sixteen hydrogens. The ratio of carbons to hydrogens is 8-to-16, which reduces to 1-to-2. The full formula for 4-octene is 
, and the empirical formula is 
.
4-octene looks like this:
To get an empirical formula, we find the ratio of each element within the compound and make it as low as possible. We have eight carbons and sixteen hydrogens. The ratio of carbons to hydrogens is 8-to-16, which reduces to 1-to-2. The full formula for 4-octene is
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What is the mass percentage of aluminum in aluminum (III) oxide?
What is the mass percentage of aluminum in aluminum (III) oxide?
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Aluminum oxide has the formula 
.
Aluminum has a molecular weight of 
, and oxygen has a weight of 
. Using these values, we can calculate the molecular weight of aluminum oxide.
The mass percentage is given by the mass of aluminum divided by the total molecular weight.
Aluminum oxide has the formula
Aluminum has a molecular weight of
The mass percentage is given by the mass of aluminum divided by the total molecular weight.
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How many grams of nitrogen are in 50g of ammonium sulfate?
How many grams of nitrogen are in 50g of ammonium sulfate?
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First convert grams of ammonium sulfate to moles, then use the mole-to-mole ratio between nitrogen and ammonium sulfate. Finally, convert moles of nitrogen back into grams.
First convert grams of ammonium sulfate to moles, then use the mole-to-mole ratio between nitrogen and ammonium sulfate. Finally, convert moles of nitrogen back into grams.
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Manganese forms a number of oxides, one of which is composed of 72% manganese by mass. Which of the following is the formula for this oxide?
Manganese forms a number of oxides, one of which is composed of 72% manganese by mass. Which of the following is the formula for this oxide?
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Given that the molar mass of oxygen is about 16g, and molar mass of manganese is about 55g, 
contains 165g of manganese and 64g of oxygen, for a total of 229g.
165g/229g = 0.72
So, the ratio of manganese to oxygen in this compound is 72% manganese by mass.
Given that the molar mass of oxygen is about 16g, and molar mass of manganese is about 55g,
165g/229g = 0.72
So, the ratio of manganese to oxygen in this compound is 72% manganese by mass.
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Convert 23g of water to moles.
Convert 23g of water to moles.
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First find the molar mass of water (H2O). You should be comfortable with the molar masses of hydrigen and oxygen from memory to reduce time on the MCAT exam.
Next, solve for moles.
First find the molar mass of water (H2O). You should be comfortable with the molar masses of hydrigen and oxygen from memory to reduce time on the MCAT exam.
Next, solve for moles.
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How many carbon atoms exist in two moles of carbon dioxide?
How many carbon atoms exist in two moles of carbon dioxide?
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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.
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
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What is the empirical formula for a compound that contains 41.4% carbon, 6.9% hydrogen, 27.6% oxygen, and 24.1% nitrogen?
What is the empirical formula for a compound that contains 41.4% carbon, 6.9% hydrogen, 27.6% oxygen, and 24.1% nitrogen?
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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.
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.
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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Which of the following answer choices is not written as an empirical formula?
Which of the following answer choices is not written as an empirical formula?
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An empirical formula must be written as the most simplified ratio of the elements that the compound contains. For example, 
is empirical because it cannot be simplified any further; the ratio of its atoms is 1:1:4.
The formula for glucose, 
, can be simplified by a factor of six. The empirical formula for glucose would be 
.
An empirical formula must be written as the most simplified ratio of the elements that the compound contains. For example,
The formula for glucose,
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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 = 
What is the empirical formula of this compound?
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
Molecular weight of this compound =
What is the empirical formula of this compound?
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The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.
The molecular weight of carbon is 
, hydrogen is 
, and oxygen is 
.
The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.
After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case, 
of oxygen).
You are left with 
carbons, 
hydrogens, and 
oxygen. The empirical formula for this compound is 
.
The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.
The molecular weight of carbon is
The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.
After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case,
You are left with
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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 =
Compared to the empirical formula, the molecular formula contains more atoms of carbon and more atoms of oxygen.
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
Molecular weight of this compound =
Compared to the empirical formula, the molecular formula contains more atoms of carbon and more atoms of oxygen.
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The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.
After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case, of oxygen). You are left with carbons, hydrogens, and oxygen. The empirical formula for this compound is .
To find the molecular formula of the compound you need to divide the molecular weight of the actual compound by the molecular weight of the empirical formula. The molecular weight of the empirical formula is:
Dividing the molecular weight of the actual compound (
) by the molecular weight of empirical formula gives:
This means that the empirical formula must be multiplied by three to get the molecular formula; therefore, the molecular formula is 
. Compared to the empirical formula, the molecular formula contains 
more carbon atoms and 
more oxygen atoms.
The first step in solving this question is to convert the mass of each element to moles. This can be done by dividing the given mass of each element by the molecular weight of each element.
After finding the moles of each element, you have to find the smallest whole number ratio of each element. The smallest whole number ratio can be found by dividing moles of each element by the lowest mole quantity (in this case,
To find the molecular formula of the compound you need to divide the molecular weight of the actual compound by the molecular weight of the empirical formula. The molecular weight of the empirical formula is:
Dividing the molecular weight of the actual compound (
This means that the empirical formula must be multiplied by three to get the molecular formula; therefore, the molecular formula is
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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?
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
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?
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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.
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 (
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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 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
Molecular weight of this compound =
The researcher obtains a sample of
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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.
A mole is defined as the number of atoms present in
Remember that a mole of any element contains
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Which of the following is both an empirical formula and a molecular formula?
Which of the following is both an empirical formula and a molecular formula?
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An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.
The only given answer that is not divisible by a whole number is 
, making it an empirical formula. It is also the molecular formula for both acetaldehyde and ethanol, depending on molecular geometry and orientation.
An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.
The only given answer that is not divisible by a whole number is
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Which of the following is a molecular formula but not an empirical formula?
Which of the following is a molecular formula but not an empirical formula?
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An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.
The only given answer that is fully divisible by a whole number is 
. This means it is not an empirical formula. Any given formula, however, will be a molecular formula, provided that the atoms are capable of forming the appropriate bonds. This is the molecular formula for glucose, which reduces to the empirical formula of 
. 
is the empirical formula for a generalized carbohydrate molecule.
An empirical formula is the simplest form of a molecular formula that still retains the ratio of the elements. If a formula can be divided by a whole number, it is a molecular formula and not an empirical formula. A molecular formula is the exact identity of a compound, showing the total number of atoms used to create the compound.
The only given answer that is fully divisible by a whole number is
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Which of the following contains the greatest number of moles?
Which of the following contains the greatest number of moles?
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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.
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.
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Which of the following cannot be directly related to Avogadro's number by stoichiometry?
Which of the following cannot be directly related to Avogadro's number by stoichiometry?
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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.
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.
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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?
Consider the following molecular formulas:
*The IUPAC name for DEET is N,N-diethyl-meta-toluamide
DEET has a density of
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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.
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.
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