1) Determine the percentage of calcium ion in Ca₃(PO₄)₂ . A)38.765% B)12.92% (C).19.97%, (D)2.5797%

2. A sample of pure copper has a mass of 16.35 g. How many moles does this mass correspond to?

A)3.887 mol (B)1.549 x 10²³ mol (C.)0.2573 mol (D.)2.715 x 10^-23 mol

3. Which sample contains the greatest number of atoms? (A)25g Cu (B)25g F₂ (C)25g Ne (D)25g Ba

4.What is the percentage by mass of O in Cu₄(AsO₃)₂(CH₃CO₂)₂? A)3.863% (B)25.885% (C)14.866% (D)6.727%

5.If a compound has a ratio of one nitrogen to two hydrogen, what is the percentage of nitrogen in the compound?

A)87.5%, B)66.7% C)33.3% D)12.5%

6.A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. Select all that is correct.

A)The empirical formula is CH.

B.)The molecular formula is C₈H₈.

C.)The molecular formula is C₆H₆.

D.)There is not enough information for us to determine the molecular and the empirical formulas.

7.A compound has the empirical formula CH₂O and a formula mass of 120.10 amu. What is the molecular formula of the compound?

A)C₄H₈O₄ (B)C₃H₆O₃ (C)C₂H₄O₂ D)CH₂O

8.Consider the following aqueous solutions. Which solute exists as a mixture of molecules and ions solution?

A)A sample of sodium chloride in water.

B)A sample of hydrogen chloride in water.

C)A sample of glucose in water.

D)A sample of acetic acid in water.

9.What is the molar concentration of sodium ions in 0.652 M sodium sulfate solution?

A)0.652 M (B)1.30 M (C)0.326 M (D)0.767 M

10.What is the molar concentration of a 250.0 mL solution that contains 0.5280 g of NaOH?

A)0.5280 M (B)18.94 M (C)0.05280 M (D)0.005280 M

11.A 250-mL sample of 1.25 M NaOH solution is prepared by dissolving an amount of 5.28 M NaOH. Select all that is correct.

A)59.2 mL of the 5.28 M NaOH was used in the dilution.

B)The moles of NaOH transferred from the 5.28 M NaOH solution are equal to the number of moles in the 250-mL sample of the NaOH solution.

C.)There are fewer moles of NaOH in the dilute solution than in the amount of the concentrated solution used.

D)1.69 x 10^-2  mL

12.How many grams of NaOH are in 500.0 mL of 2.65 M NaOH solution?

A)53.0 g (B)5.30 x 10² g (C)0.0189 g (D)1.33 g

 

Answer:

1. A)38.765% 
2. (C.)0.2573 mol 
3. (C)25g Ne
4. (B)25.885% 
5. A)87.5%
6. A)The empirical formula is CH. and C.)The molecular formula is C₆H₆.
7. A)C₄H₈O₄ 
8. D)A sample of acetic acid in water.
9. (B)1.30 M 
10. (C)0.05280 M 
11. A)59.2 mL of the 5.28 M NaOH was used in the dilution. and B)The moles of NaOH transferred from the 5.28 M NaOH solution are equal to the number of moles in the 250-mL sample of the NaOH solution.
12. A)53.0 g 

Step-by-step explanation

1. First calculate the molecular mass of Ca₃(PO₄)₂ using the atomic mass of each element from the periodic table. Here, the molecular mass is equal to the sum of product of the atomic mass of each element and their corresponding subscript.

molecular mass = 3 x atomic mass of Ca + 2 x atomic mass of P + 8 x atomic mass of O

molecular mass = 3 x 40.08 amu + 2 x 30.97 amu + 8 x 16.00 amu

molecular mass = 310.18 amu

Then, calculate the total mass of the Ca²⁺ (calcium ion) component by multiplying its atomic mass and subscript as shown below.

total mass of Ca²⁺ = 3 x atomic mass of Ca

total mass of Ca²⁺ = 3 x 40.08 amu

total mass of Ca²⁺ = 120.24 amu

Finally, calculate the percentage calcium ion using the formula below.

percentage calcium ion = (total mass of Ca²⁺ / molecular mass) x 100% = (120.24 amu / 310.18 amu) x 100% = 38.765% (A)

2. Calculate the number of moles of 16.35 g pure copper (Cu) from its molar mass from the periodic table which is 63.546 g/mol using the formula below.

number of moles = mass / molar mass = (16.35 g) / (63.546 g/mol) = 0.2573 mol (C)

3. From the formula below, the number of moles is inversely proportional to the molar mass.

number of moles = mass / molar mass

Now, from the molar masses of the choices which were searched from the internet/database, Ne (20.18 g/mol) has the lowest molar mass compared to Cu (63.55 g/mol), F₂(38.00 g/mol), and Ba (137.33 g/mol), hence, 25 g Ne will have the highest number of moles. Note, that the number of moles is directly proportional to the number of atoms, hence, 25 g Ne (C) will also have the greatest number of atoms. 

4. First calculate the molecular mass of Cu₄(AsO₃)₂(CH₃CO₂)₂ using the atomic mass of each element from the periodic table. Here, the molecular mass is equal to the sum of product of the atomic mass of each element and their corresponding subscript.

*molecular mass = 4 x atomic mass of Cu + 2 x atomic mass of As + 10 x atomic mass of O + 4 x atomic mass of C + 6 x atomic mass of H

molecular mass = 4 x 63.55 amu + 2 x 74.92 amu + 10 x 16.00 amu + 4 x 12.01 amu + 6 x atomic mass of 1.01 amu

molecular mass = 618.11 amu

Then, calculate the total mass of the O component by multiplying its atomic mass and subscript as shown below.

total mass of O = 10 x atomic mass of O

total mass of O = 10 x 16.00 amu

total mass of O = 160.0 amu

Finally, calculate the percentage by mass of O  using the formula below.

percentage by mass of O = (total mass of O / molecular mass) x 100% = (160.0 amu / 618.11 amu) x 100% = 25.885% (B)

5. Since compound has a ratio of one nitrogen to two hydrogen, then, its empirical formula is NH₂. Then, calculate the molecular mass of NH₂ using the atomic mass of each element from the periodic table. Here, the molecular mass is equal to the sum of product of the atomic mass of each element and their corresponding subscript.

molecular mass = 1 x atomic mass of N + 2 x atomic mass of H

molecular mass = 1 x 14.0 amu + 2 x 1.0 amu

molecular mass = 16.0 amu

Then, calculate the total mass of the N (calcium ion) component by multiplying its atomic mass and subscript as shown below.

total mass of N = 1 x atomic mass of N

total mass of N = 1 x 14.0 amu

total mass of N = 14.0 amu

Finally, calculate the percentage N using the formula below.

percentage N = (total mass of N / molecular mass) x 100% = (14.0 amu / 16.0 amu) x 100% = 87.5% (A)

6. First, calculate the percentage by mass of H using the formula below.

100% = percentage by mass of H + percentage by mass of C

100% = percentage by mass of H + 92.3% 

100% - 92.3% = percentage by mass of H

percentage by mass of H = 7.7%

Calculate the total mass of C and H in the empirical formula using the formula below.

total mass = molar mass x (percentage by mass / 100%)

total mass of C = 78.1 g/mol x (92.3% / 100%) = 72.09 g/mol

total mass of H = 78.1 g/mol x (7.7% / 100%) = 6.01 g/mol

Then, calculate the number of C and H in the molecular formula using the formula below.

number of element = total mass / molar mass

number of C = (72.09 g/mol) / (12.01 g/mol) = 6 C

number of H = (6.01 g/mol) / (6.01 g/mol) = 6 H

Hence, the molecular formula of the compound is C₆H₆ (C.). Now, if we divide the subscript of all the elements by the common factor which is 6, we will get the empirical formula which is CH (A.).

7. First calculate the  mass of CH₂O  using the atomic mass of each element from the periodic table. Here, the molecular mass is equal to the sum of product of the atomic mass of each element and their corresponding subscript.

 mass = 1 x atomic mass of C + 2 x atomic mass of H + 1 x atomic mass of O

 mass = 1 x 12.01 amu + 2 x 1.01 amu + 1 x 16.00 amu

 mass = 30.03 amu

Then, determine the ratio of the formula mass and the mass of the empirical formula as shown below.

ratio of formula mass and the mass of the empirical formula  = 120.10 amu / 30.03 amu = 4

This ratio is the ratio of the number of each atom in the  molecular to the empirical formula. Hence, the empirical formula can be determined by multiplying the subscript in CH₂O by 4 which will result to C₄H₈O₄ (A).

8. To answer this let's examine each choices.

• A. Sodium chloride is a strong electrolyte, hence, it will ionized completely into its ions, leaving no sodium chloride molecule. Hence, A is not the answer.
• B. Hydrogen chloride is a strong acid and a strong electrolyte, hence, it will ionized completely into its ions, leaving no hydrogen chloride molecule. Hence, B is not the answer.
• C. Glucose is a non-electrolyte, hence, it will not form ions. Hence, C is not the answer.
• D. Acetic acid is a weak acid and a weak electrolyte, hence, some of the molecular weak acid will ionized into its ions. At equilibrium, there will be molecular acetic acid and ions, hence, D is the answer.

9. Note that there are 2 sodium ions (Na²⁺) per 1 sodium sulfate (Na₂SO₄), hence, the concentration of sodium ion is twice the concentration of sodium sulfate as shown below.

concentration of sodium ions = 0.652 M sodium sulfate (2 sodium ions / 1 sodium sulfate) = 1.30 M sodium ions (B)

10.  First, calculate the number of moles of 0.5280 g NaOH from its molar mass from the internet/database which is 40.00 g/mol using the formula below.

number of moles = mass / molar mass = (0.5280 g) / (40.00 g/mol) = 0.0132 mol NaOH

Then, convert the volume to L using the conversion factor: 1 L / 1000 mL. Note that this is done so that the unit of volume is consistent with the unit of molar concentration.

Volume (in L) = 250.0 mL x (1 L / 1000 mL) = 0.2500 L

Then, calculate the molar concentration using the formula below.

molar concentration = number of moles / Volume = (0.0132 mol) / 0.2500 L = 0.05280 M NaOH (C)

11. Calculate the volume of 5.28 M NaOH needed to prepare 250 mL of 1.25 M NaOH using the formula below.

V_stock = V_standard (M_standard / M_stock)

Where V_stock  = volume of stock solution to be added, V_standard = total volume of the new standard solution, M_standard = concentration of the new standard solution, and M_stock = concentration of the stock solution being added.

V_stock = 250 mL x (1.25 M/ 5.28 M)

V_stock = 59.2 mL (A)

Hence, statement A is true. Note that in dilution, the number of moles in the added stock solution is equal and will be the number of moles in the final standard solution since the stock solution is the only source of the solute, hence, B is true. C is wrong as it contradicts the previous statement. Lastly, D is wrong as the volume needed have already been calculated to be 59.2 mL.

12. First, convert the volume to L using the conversion factor: 1 L / 1000 mL. Note that this is done so that the unit of volume is consistent with the unit of molar concentration.

Volume (in L) = 500. mL x (1 L / 1000 mL) = 0.5000 L

Then, calculate the number of moles using the formula below.

number of moles = molar concentration x volume = 2.65 M x 0.5000 L = 1.325 mol NaOH

Finally, calculate the mass in 1.325 mol NaOH from its molar mass from the internet/database which is 40.00 g/mol using the formula below.

mass = number of moles x molar mass = 1.325 mol x 40.00 g/mol = 53.0 g (A)