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AP Chemistry Composition of Substances and Solutions — practice questions

Explore 23 free Composition of Substances and Solutions questions in AP Chemistry Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Composition of Substances and Solutions questions and exercises with answers

Showing questions 1–23 of 23.

To count atoms easily, how can the formula of calcium phosphate, Ca3(PO4)2, be rewritten in a simpler format?Read the answer and worked solution →Calcium phosphate, Ca3(PO4)2, has a formula mass of 310.18 amu. What is the mass of 0.500 mol of calcium phosphate?Read the answer and worked solution →What is the mass of 0.443 mol of hydrazine, N2H4? Use N = 14.007 and H = 1.008 g/mol.Read the answer and worked solution →A prospector collects 15.00 g of pure gold. Given that 2.561 mol of gold has a mass of 504.4 g, how many Au atoms are in the sample?Read the answer and worked solution →A 55 kg woman has 7.5 x 10^-3 mol of hemoglobin, molar mass 64,456 g/mol, in her blood. What mass of hemoglobin is that?Read the answer and worked solution →Aluminium sulfate, Al2(SO4)3, has a formula mass of 342.14 amu. How many moles are in 78.452 g of it?Read the answer and worked solution →A compound sample contains 1.71 g of carbon and 0.287 g of hydrogen. What is its empirical formula? Use C = 12.01 and H = 1.008 g/mol.Read the answer and worked solution →Tea sweetened to a sucrose molarity of 0.05 M is poured out. What volume of the tea holds 0.004 mol of sucrose?Read the answer and worked solution →What volume of a 1.50 M KBr solution contains 0.555 mol of KBr?Read the answer and worked solution →What is the molarity of a solution holding 7.0 x 10^-3 mol of iodine, I2, in a total volume of 100.0 mL?Read the answer and worked solution →A solution contains 0.444 mol of CoCl2 in 0.654 L. What is its molarity?Read the answer and worked solution →A 0.150 L sample of 0.556 M NaCl is left to evaporate until only 0.105 L remains. What is the new concentration?Read the answer and worked solution →A laboratory needs 250 mL of 2.00 M HCl. How many millilitres of 11.9 M HCl must be diluted to prepare it?Read the answer and worked solution →What volume of a 1.00 M Fe(NO3)3 stock can be diluted to prepare 1.00 L of a 0.250 M solution?Read the answer and worked solution →If 4.12 L of 0.850 M H3PO4 is diluted to a final volume of 10.00 L, what is the resulting concentration?Read the answer and worked solution →Liquid bleach is 7.4% sodium hypochlorite by mass. What mass of NaOCl is in a 250.0 g bottle of the bleach?Read the answer and worked solution →The EPA maximum safe level of fluoride ion in tap water is 4 ppm. At that level, what mass of fluoride is in 1.00 L of tap water, taking its density as 1.00 g/mL?Read the answer and worked solution →Aspirin is C9H8O4 with a molar mass of 180.159 g/mol. Using H = 1.008 g/mol, what is its mass percentage of hydrogen?Read the answer and worked solution →What is the mass of 0.600 mol of ozone molecules, O3? Use O = 16.00 g/mol.Read the answer and worked solution →In ammonia, NH3, the three hydrogen atoms together weigh 3.024 amu and the formula mass is 17.03 amu. What is the mass percentage of hydrogen?Read the answer and worked solution →Crunch the mole numbersRead the answer and worked solution →Pick the heavier moleRead the answer and worked solution →Rebuild the formula huntRead the answer and worked solution →

Worked practice questions

To count atoms easily, how can the formula of calcium phosphate, Ca3(PO4)2, be rewritten in a simpler format?

  1. Ca3PO4
  2. Ca3P2O8
  3. Ca3P2O12
  4. Ca2P3O8
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Correct answer: B. Ca3P2O8

1. The bracket subscript 2 multiplies everything inside the parentheses. 2. Two phosphate groups give 2 phosphorus atoms and 2 x 4 = 8 oxygen atoms. 3. With the 3 calcium atoms the simple format is Ca3P2O8. 4. Reading the bracket contents only once gives Ca3PO4, which loses a whole phosphate group. _Source: OpenStax Chemistry (CC BY 4.0), Ch 3 "Composition of Substances and Solutions", section 3.1 Formula Mass and the Mole Concept_

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Calcium phosphate, Ca3(PO4)2, has a formula mass of 310.18 amu. What is the mass of 0.500 mol of calcium phosphate?

  1. 31.02 g
  2. 310.2 g
  3. 620.4 g
  4. 155.1 g
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Correct answer: D. 155.1 g

1. A formula mass of 310.18 amu means a molar mass of 310.18 g/mol, since the two are numerically equal. 2. Mass equals moles multiplied by molar mass. 3. Substituting gives 0.500 mol x 310.18 g/mol = 155.1 g. 4. Units check: mol x g/mol leaves grams. 5. The value 310.2 g ignores that only half a mole is present, and 620.4 g divides by 0.500 instead of multiplying. _Source: OpenStax Chemistry (CC BY 4.0), Ch 3 "Composition of Substances and Solutions", section 3.1 Formula Mass and the Mole Concept_

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What is the mass of 0.443 mol of hydrazine, N2H4? Use N = 14.007 and H = 1.008 g/mol.

  1. 14.2 g
  2. 12.4 g
  3. 1.42 g
  4. 0.0138 g
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Correct answer: A. 14.2 g

1. Build the molar mass first: nitrogen contributes 2 x 14.007 = 28.014 g/mol. 2. Hydrogen contributes 4 x 1.008 = 4.032 g/mol, so the molar mass is 32.05 g/mol. 3. Mass equals moles times molar mass: 0.443 mol x 32.05 g/mol = 14.2 g. 4. The value 12.4 g uses only the nitrogen part of the molar mass, dropping the four hydrogen atoms. 5. The value 0.0138 g divides the moles by the molar mass instead of multiplying. _Source: OpenStax Chemistry (CC BY 4.0), Ch 3 "Composition of Substances and Solutions", section 3.1 Formula Mass and the Mole Concept_

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