Moles and mass are related by molar mass. Use molar mass as the conversion factor
grams → moles; Conversion factor: 1mol = 42.08 g.
Calculation: 25.0g × (1 mol / 42.08 g) = 0.598 mol. Always check the units in the calculation. You can see that grams cancel to give moles.
Check: The molar mass of propylene tells us that 42.08 g is the mass of 1 mole of propylene. Therefore 25 grams should be less than a mole.
Moles and mass are related by molar mass. Use molar mass as the conversion factor.
Moles and mass are related by molar mass. Use molar mass as the conversion factor.
Moles and mass are related by molar mass. Use molar mass as the conversion factor.
Check your calculation. You may not have entered the calculation correctly in the calculator. All values multiplied together in the denominator must be enclosed in a single parenthesis.
4 x 1012 molecules is less than 1 mole!
molecules → moles; Conversion factor: 1 mol = 6.022 ×1023
Calculation: 5.4 x 1012 molecules × (1 mol / 6.022 ×1023 molecules) = 9.0 x 10-12 after rounding to two significant digits.
Note that when inputting the calculation as a single line in a calculator, a parenthesis must be used as follows: 5.4 x 1012 ÷ (6.022 × 1023)
A collection of 6.022 × 1023 urea molecules is 1 mole of urea.
4 x 1012 molecules should be less than 1 mole!
A collection of 6.022 × 1023 urea molecules is 1 mole of urea. Use this information as a conversion factor.
A collection of 6.022 × 1023 urea molecules is 1 mole of urea. Use this information as a conversion factor.
Convert number of molecules to grams via moles. molecules → moles → grams.
Moles and molecules are related by the Avogadro’s number. Moles and mass in grams are related by the molar mass.
Convert number of molecules to grams via moles. molecules → moles → grams.
Moles and molecules are related by the Avogadro’s number. Moles and mass in grams are related by the molar mass.
Convert number of molecules to grams via moles. molecules → moles → grams.
Moles and molecules are related by the Avogadro’s number. Moles and mass in grams are related by the molar mass.
You may have inverted the conversion factor between moles and molecules.
1 mol = 6.022 × 1023 Write the conversion factor so that units in the denominator cancel out the units in the numerator.
molecules → moles → grams.
2 molecules × ( 1 mol / 022×1023 molecules) × (18.02 g/ 1 mol) = 6×10-23 g after rounding to one significant digit.
You can see that the units in the numerator and denominator cancel to give grams as the final unit.
lb → grams → moles → molecules. Write conversion factors so that units in the numerator and denominator cancel.
25 lb × (454 g/ 1 lb ) × (1 mol / 283.28 g) × (6.022×1023 molecules / 1 mol) = 2.4 ×1025.
Convert the given mass to grams. Then grams can be converted to molecules via moles
You may not have converted lb to grams.
You may not have used parenthesis while inputting the numbers into the calculator. Note that all values multiplied together in the denominator must be enclosed in a single parenthesis.
You may have inverted the conversion factor between grams and moles.
1 mol = 283.28 g. Write the conversion factor so that units in the denominator cancel out the units in the numerator.
A collection of 5.60× 1022 molecules is 0.093 mol, which corresponds to 16.8 g.
Strategy: 5.60 × 1022 molecules → moles → grams
Conversion factors: 1 mol = 6.022 ×1023 molecules; 1 mol glucose = 180.16 g. Conversion factors are written so that units in the denominator cancel out the units in the numerator
molecules → moles → grams.
To find the mass of 5.60 × 1022 molecules, first find how many moles 5.60 × 1022 molecules corresponds to. We know that 022×1023 molecules correspond to 1 mole.
molecules → moles → grams.
To find the mass of 5.60 × 1022 molecules, first find how many moles 5.60 × 1022 molecules corresponds to. We know that 6.022×1023 molecules correspond to 1 mole.
You may have inverted the conversion factor between grams and moles.
1 mol of glucose = 180.16 g. Write the conversion factor so that units in the denominator cancel out the units in the numerator.
You may have thought that what is given is moles. Note that what is given is molecules.
60 × 1022 molecules is less than one mole. So, the mass should be less than the molar mass
What you have calculated is moles of Na2SO4. However, what is asked is moles of Na. Convert moles of Na2SO4 to moles of Na using the molar ratio between Na2SO4 and Na.
Strategy:4 g Na2SO4 → mol Na2SO4 → mol of Na
Conversion factors: Use the molar mass of Na2SO4 to convert grams to moles. Then, use the molar ratio between Na2SO4 and Na (since 1 Na2SO4 unit has 2 Na atoms, 1 mole Na2SO4 units will have 2 mole Na atoms. The molar ratio is 1 mol Na2SO4 = 2 mol Na)
Calculation: 5.4 g Na2SO4 × (1 mol Na2SO4 / 142.04 g Na2SO4) × ( 2 mol Na / 1 mol Na2SO4).
Notice that the units in the denominator cancel out the units in the numerator to yield mol of Na as the final unit.
You may have inverted the molar ratio between Na2SO4 and Na.
1 mol Na2SO4 = 2 mol Na. Write conversion factor so that the unit in the denominator cancels the unit in the numerator.
You may have used mass of sodium in the denominator! Note that what you are given is 5.4 g of Na2SO4. To cancel grams of Na2SO4 conversion factor should have grams of Na2SO4 in the denominator.
Hint: The problem involves two conversion factors: molar mass of Na2SO4, molar ratio between Na2SO4 and Na.
You may have used molar mass of sodium in the denominator! Note that what you are given is 5.4 g of Na2SO4. To cancel grams of Na2SO4 , the first conversion factor should have grams of Na2SO4 in the denominator.
Hint: The problem involves two conversion factors: molar mass of Na2SO4, molar ratio between Na2SO4 and Na.
You may have counted oxygen atoms twice!
Strategy: g of Ca(NO
3)
2 →
mol Ca(NO3)2 → mol of O → atoms of O
Conversion factor for
mol Ca(NO3)2 → mol of O is 1 mol Ca(NO
3)
2 = 6 mol O (1:6 molar ratio)
OR,
g of Ca(NO
3)
2 → mol Ca(NO
3)
2 →
molecules of Ca(NO3)2 → atoms of O
Conversion factor for
molecules of Ca(NO3)2 → atoms of O is 1 Ca(NO
3) molecule = 6 oxygen atoms (1:6 ratio).
Note: ‘formula unit’ is the correct terminology instead of ‘molecule’ when referring to ionic compounds. For simplicity, the word molecule was used here.
Review similar problem in worked out
example 3.8
You may have used mass of oxygen in the denominator! Note that what you are given is 1.5 g of Ca(NO3)2. To cancel grams of Ca(NO3)2, the first conversion factor should have grams of Ca(NO3)2 in the denominator.
Hint: If you know how many Ca(NO3)2 units there are in 1.5 g Ca(NO3)2, you can count the oxygen atoms they contain-- knowing that each Ca(NO3)2 unit contains 6 oxygens.
Review similar problem in worked out example 3.8
5× 1021 is how many Ca(NO3)2 units there are, NOT how many oxygens! Note that each Ca(NO3)2 unit has 6 oxygens
You may have inverted the conversion factor involving Avogadro’s number. Note that 9.1×10-26 has a negative exponent, so, this value is less than 1! The sample 1.5 g sample should contain lots of oxygen atoms.
Sorry, 3.89 × 10
27 is not the correct answer! The formula for carbon tetrachloride is CCl
4. Review similar problem in worked out
example 3.8.
Sorry, 9.73 × 10
26 is not the correct answer! The formula for carbon tetrachloride is CCl
4. Review similar problem in worked out
example 3.8.
Strategy:
g of CCl4 → mol of CCl4 → molecules of CCl4 → atoms of Cl
(Note: 1:4 ratio is used in the third conversion factor).
OR,
g of CCl4 → mol of CCl4 → mol of Cl → molecules of Cl
(Note: 1:4 molar ratio is used in the second conversion factor).
The formula for carbon tetrachloride is CCl
4. You may not have considered that one CCl
4 molecule has 4 chlorine atoms. Review similar problem in worked out
example 3.8.
Sorry, 1.13 × 10
21 is not the correct answer! The formula for carbon tetrachloride is CCl
4. Review similar problem in worked out
example 3.8.
Error in counting ions; you may have counted 17 in each formula unit! Note that there are two aluminum ions and three sulfate ions, that is 2+3=5 ions, in each Al2(SO4)2 unit (note that sulfate is a polyatomic ion, don’ dissociate it into sulfurs and oxygens!)
What you calculated (ie. 3.3 × 1024) is how many Al2(SO4)3 units there are. You did not consider that each Al2(SO4)3 is made of positive and negative ions. Report how many ions there would be in 3.3 × 1024 Al2(SO4)3 units.
You may have counted only Aluminum ions. Count both Aluminum and sulfate ions. Note that there are two aluminum ions and three sulfate ions, that is 2+3=5 ions, in each Al2(SO4)2 units.
There are 5 ions in each Al2(SO4)3 unit, ie. 1:5 ratio between Al2(SO4)3 and ions.
Strategy: Step-1: 4 moles Al2(SO4)3 → number of molecules of Al2(SO4)3 using Avogadro’s number as conversion factor; Step-2: molecules of Al2(SO4)3 → number of ions using 1:5 ratio.
OR,
Step-1: 5.4 moles Al2(SO4)3 → mol of ions using 1:5 molar ratio; Step-2: mol of ions → number of ions using Avogadro’s number.
You may have counted only sulfate ions. Count both Aluminum and sulfate ions. Note that there are two aluminum ions and three sulfate ions, that is 2+3=5 ions, in each Al2(SO4)2
2 mol of C2H5OH contains 2 mol of oxygen (O) equivalent to 32 grams. There is another substance in the options that contains more than 32 grams.
1 mol of HCO2H contains 2 mol of oxygen (O) equivalent to 32 grams. There is another substance in the options that contains more than 32 grams.
1 mol of H2O contains 1 mol of oxygen (O) equivalent to 16 grams. There is another substance in the options that contains more than 16 grams.
1 mol of SO3 contains 3 mol of oxygen (O) equivalent to 48 grams. Other substances in the options contain either 16 g or 32 g of oxygen.
1 mol of CO2 contains 2 mol of oxygen (O) equivalent to 32 grams. There is another substance in the options that contains more than 32 grams.
The value, 94.6 g, is too much!
1 NH3 has 3H. Interpret this in terms of mass and use the mass relationship as a conversion factor.
Strategy: 5.60 g of hydrogen (H) → grams of NH3.
1 mole of NH3 contains 3 moles of H. We can interpret this in terms of mass and say 17.02 g of NH3 contains 3×008 g H (which equals 3.02 g ).
So, the conversion factor will be: 17.02 g NH3 = 3.02 g H.
Calculation: 5.6 g H (17.02 g NH3 / 3.02 g H) = 31.5 g H.
Note that the unit in the denominator cancels the unit in the numerator.
1 NH3 has 3H. Interpret this in terms of mass and use the mass relationship as a conversion factor.
1 NH3 has 3H. Interpret this in terms of mass and use the mass relationship as a conversion factor.
1 NH3 has 3H. Interpret this in terms of mass and use the mass relationship as a conversion factor.
Percent composition of sulfur is percentage of sulfur “by mass” in the compound.
Percent is (part / whole)× 100, part being Sulfurs mass, whole being the compound’s mass.
1 mole Na2S2O3 has 2 moles of S. Interpret this in terms of mass and calculate the percentage.
You may not have considered that there are two sulfur atoms in each Na2S2O3 Both sulfur atoms contribute to the mass percentage of sulfur.
That is not the correct answer. You may have not considered that the percentage should be percentage by “mass”.
Percent is (part / whole)× 100, part being Sulfurs mass, whole being the compound’s mass.
Percent composition of sulfur is percentage of sulfur “by mass” in the compound.
Percent is (part / whole)× 100, part being Sulfurs mass, whole being the compound’s mass.
Mass percent of sulfur = ((2 x 32 g) /11 g) × 100
The value 6.73 is too small! Note that percent composition means percentage by “mass”.
Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.
Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.
Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.
= ((3× 40.08 g) / 210.18 g) × 100 = 39%
(Note: Do not get confused because the problem states calcium “ion”. Ions weigh the same as atoms because removing or adding electrons has negligible effect on the mass of an atom. The problem should be solved as you would normally solve a percent composition problem.)
Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.
You may not have considered that there are 3 calcium ions in each Ca3(PO4)2. All the three calcium ions contribute to the mass percentage of calcium.
You may have calculated simplest ratio of grams, which is not correct. Note that empirical formula is the simplest whole number ratio of
moles!
Review similar worked out problem in
example 3.12.
ote that C
2H
2Cl
4 can be simplified to CHCl
2. So, C
2H
2Cl
4 cannot be the empirical formula! Also, the ratio is not correct.
Review similar worked out problem in
example 3.12.
It is true that the ratio of moles is 2:4:2. However, this is not the simplest ratio!
Empirical formula is the simplest whole number ratio of moles.
Review similar worked out problem in
example 3.12.
Assuming 100 g of the compound:
Ratio of masses: 3 g C, 4.1 g H, 71.6 g Cl
Ratio of moles: 2.023 mol C, 4.067 mol H, 2.020 mol Cl
This results in a relative ratio of 1.001 : 2.013 : 1.000, which is close to whole number ratio, 1:2:1. Hence, the Empirical formula is CH
2Cl.
Review similar worked out problem in
example 3.12.
Do not ignore chlorine in the compound! Since mass percentages of all components should add up to 100%, Chlorine’s mass percentage will be 71.6%.
Review similar worked out problem in
example 3.12.
This is not the correct answer. If you have calculated the percentages from the formula, you can see that the method may not always work. Determine the empirical and molecular formulas systematically as shown in a similar worked out problem,
example 3.12.
Grams: 28.03 g Mg, 21.60 g Si, 1.16 g H, 49.21 g O (assuming 100 g sample)
Moles: 1.17 mol Mg, 0.769 mol Si, 1.15 mol H, 3.08 g O. This results in a relative ratio of 1.5 : 1 : 1.5 : 4; doubling each results in simplest
whole number ratio, 3:2:3:8. Therefore, empirical formula is Mg
3Si
2H
3O
8. The molar mass is twice the empirical formula mass. Hence, the molecular formula is Mg
6Si
4H
6O
16.
Review similar worked out problem in
example 3.12.
Unless simplest ratio is very close to a whole number ratio, do not round!
Review similar worked out problem in
example 3.12.
Grams: 40.0 g C, 6.71 g H, 53.28 g O.
Moles: 3.331 mol C, 6.657 mol H, 3.330 mol O.
This results in a relative ratio of 1.000 : 1.999 : 000, which is close to simplest whole number ratio of 1:2:1. Therefore, empirical formula is CH
2O.
Review similar worked out problem in
example 3.12.
M1V1 = M2V2, where, V1 and V2 are volumes of solution before and after adding water, respectively. We know the volume after dilution to be 250 mL. Calculate the volume before dilution. The difference between the two volumes is how much water is needed.
15M × V1 = 0.025M × 250 mL
V1 = 41.7 mL.
Vwater = V2 – V1 = 208 mL = 210 mL after rounding to two significant digits.
This is a dilution problem.
The solution is prepared from a 0.15 M That means, molarity of the solution before dilution, M1, is 0.150M. The molarity and volume of the solution after dilution (i.e. M2 and V2) are 0.025M and 250 mL. Using this information, we can first calculate M1 and find the volume of water.
Use the dilution equation, M1V1 = M2V2. Note that V1 and V2 are volumes of solution before and after adding water.
Looks like you stopped after calculating the volume of solution before dilution. The volume of water needed is given by the difference in the volumes before and after dilution.
Use the dilution equation, M1V1 = M2V2. Note that V1 and V2 are volumes of solution before and after adding water.
Rearranging the molarity equation gives:
moles = molarity × Liters = 0.0120
M ×02315
L = 2.78 × 10
-4 mol. Check calculation from units: Note that the unit, “M” stands for “mol/L”. You can see that in the above calculation, Liters cancel out and the resulting unit is
mol.
See more example problems based on the molarity equation
here (examples 3.14 to 3.18).
You did not rearrange the molarity equation correctly! Also, pay attention to units in your calculation.
You did not convert milliliters to Liters! Recall that Molarity = mol/L.
Check if you rearranged the molarity equation correctly!
Check if you rearranged the molarity equation correctly!
This is a dilution problem. As more solvent is added, concentration decreases. We can solve for the new concentration using the dilution equation is M1V1 = M2V2. The volume and molarity before dilution, V1 and M1, are 25 mL and 0.5 M. The volume after dilution, V2, is 125 mL (since 100 mL water is added to 25 mL solution).
Calculation:500M × 25.0 mL = M2 × 125 mL. M2 = 0.500M × 25.0 mL / 125 mL = 0.100 M.
Note that volume increased by a factor of 5 (from 25 mL to 125 mL), so, concentration decreased by a factor of 5 (from 0.5 M to 0.1 M)
Sorry, 0.125
M is too low! Review the concept
here.
Check values for volume in your calculation! Volume of the solution after dilution will be 25 mL plus 100 mL.
Review the dilution concept
here.
Use the dilution equation. Note that the volumes V
1 and V
2 refer to volume before and after dilution, respectively. Review the concept
here!
Use the dilution equation. Note that the volumes V
1 and V
2 refer to volume before and after dilution, respectively. Review the concept
here!
Concentration should decrease after adding water! 2.00
M is a higher concentration than the starting concentration of 0.500
M. Hence, 2.00
M cannot be the answer.
Review the dilution concept
here.
Hint: 250 mL will not be used anywhere in the calculation. The value, 250 mL, is simply stating how much of 2.0 M HNO3 solution is available.
Hint: 250 mL will not be used anywhere in the calculation. The value, 250 mL, is simply stating how much of 2.0 M HNO3 solution is available.
Hint: 250 mL will not be used anywhere in the calculation. The value, 250 mL, is simply stating how much of 2.0 M HNO3 solution is available.
M1V1 = M2V
2.0M × V1 = 0.041M × 150 mL -----> V1 = 3.1 mL. You will need 3.1 mL of 2.0 M solution.
Note: If water is added to the 3.1 mL until the volume is 150 mL, the new molarity of the solution will be 0.041 M.
Hint: 250 mL will not be used anywhere in the calculation. The value, 250 mL, is simply stating how much of 2.0 M HNO3 solution is available.
The question can be simplified to “what volume of 2.0M solution is needed to prepare a 150 mL solution of 0.041M
M1V1 = M2V2, where, V1 and V2 are volumes of solution before and after adding water, respectively. We know the volume after dilution to be 250 mL. Calculate the volume before dilution. The difference between the two volumes is how much water is needed.
15M × V1 = 0.025M × 250 mL
V1 = 41.7 mL.
Vwater = V2 – V1 = 208 mL = 210 mL after rounding to two significant digits.
This is a dilution problem.
The solution is prepared from a 0.15 M That means, molarity of the solution before dilution, M1, is 0.150M. The molarity and volume of the solution after dilution (i.e. M2 and V2) are 0.025M and 250 mL. Using this information, we can first calculate M1 and find the volume of water.
Use the dilution equation, M1V1 = M2V2. Note that V1 and V2 are volumes of solution before and after adding water.
Looks like you stopped after calculating the volume of solution before dilution. The volume of water needed is given by the difference in the volumes before and after dilution.
Use the dilution equation, M1V1 = M2V2. Note that V1 and V2 are volumes of solution before and after adding water.
Note that molarity is moles/L. Relate moles of CaCl2 to moles of Cl-.
Note that molarity is moles/L. Relate moles of CaCl2 to moles of Cl-.
Sorry! 0.13 mol/L is the concentration of CaCl
2,
not the concentration of Cl
--
Convert molarity of CaCl2 to molarity of Cl- using the molar ratio between CaCl2 and Cl-.
Good job! CaCl
2 dissociates in water to produce one Ca
2+ and two Cl
- Note that the molar ratio between CaCl
2 and Cl
- is 1:2.
1 grams of CaCl2 in 150 mL solution. We can calculate the concentration (ie. molarity) of CaCl2 using the molarity equation. Since each CaCl2 produces two chloride ions when it dissolves in water, concentration of Chloride (Cl-) will be two times the concentration of CaCl2.
1 g CaCl2 × (1 mol CaCl2 / 110.98 g CaCl2) = 0.0189 mol CaCl2.
Molarity of CaCl2= 0.0189 mol CaCl2 / 0.15 L solution = 0.126 mol CaCl2 / L
(0.126 mol CaCl2 / L) × (2 mol Cl- / 1 mol CaCl2) = 0.25 mol Cl-/L
Note that molarity is moles/L. Relate moles of CaCl2 to moles of Cl-.
Molarity of Mg(NO3)2 x Liters of solution = moles of Mg(NO3)2 moles of Mg(NO3)2 → moles of NO3- using the molar ratio between Mg(NO3)2 and NO3-
You did not convert moles of Mg(NO3)2 to moles of NO3-
We can calculate moles of molarity and volume are given. Using the molar ratio between
Mg(NO3)2 and NO3- we can convert moles of Mg(NO3)2 to moles of NO3-.
Recall the molarity equation. Think about what is given and what you need to calculate.
Recall the molarity equation. Think about what is given and what you need to calculate.
Determine the formula for aluminum bromide. Note that we can convert bromide concentration (Br-) to AlBr3 concentration using the molar ratio.
You did not convert moles of Br- to moles of AlBr3! Note that each AlBr3 contains 3 Br- ions.
Aluminum bromide is AlBr
3.
Molarity of Br- × Liters of solution= moles of Br- .
moles of Br- → moles of AlBr3 (using the molar ratio: 1 mol AlBr3 = 3 mol Br- as the conversion factor).
mol AlBr3 → grams of AlBr3
Determine the formula for aluminum bromide. Note that we can convert bromide concentration (Br-) to AlBr3 concentration using the molar ratio.
Determine the formula for aluminum bromide. Note that we can convert bromide concentration (Br-) to AlBr3 concentration using the molar ratio.
Think about what you can calculate from Molarity of Na2SO4 and it’s volume.
Molarity of Na2SO4 × Liters of solution = moles of Na2SO4 → moles of Na → grams of Na
Using molarity of Na2SO4 and the volume of the solution, we can find moles of Na2SO4. moles of Na2SO4 can be converted to moles of sodium (Na)
Using molarity of Na2SO4 and the volume of the solution, we can find moles of Na2SO4. moles of Na2SO4 can be converted to moles of sodium (Na)
Using molarity of Na2SO4 and the volume of the solution, we can find moles of Na2SO4. moles of Na2SO4 can be converted to moles of sodium (Na)
Chloride concentration in 0.1 M NaCl solution will also be 0.1 M.
Chloride concentration in 0.1 M CaCl2 solution will be 0.2 M.
1 mole AlCl3 contains 3 moles Cl-. Due to this molar ratio, chloride concentration in 0.1 M AlCl3 solution will be 0.3 M.
NaClO3 has Na+ and ClO3- It does not contain chloride (Cl-) ions. Chloride concentration in NaClO3 solution is zero!
Though all solutions contain the same moles of solute per liter, each solute has different number of Chloride ions!