Regents Examination in Physical Setting / Chemistry: Using the Equations to Solve Chemistry Problems

Using the Equations to Solve Chemistry Problems

Important Formulas/Equations
In this guide, various problems are solved using the formulas and equations found in Reference Table.
 

1. Density:    

Problem
Calculate the mass of a sample of the element niobium (Atomic Number = 41) if the volume of the sample is 13.00 cm3.
Solution
Use Reference Table to find the density of niobium (8.570 g/cm3).
Rearrange the variables in the equation in order to solve for mass:
.
Substitute the data in the equation and perform the calculation:

 

2. Mole Calculations:    

Problem
Calculate the number of moles in a sample of oxygen gas (O2, gram-formula mass = 32.0 g/mol) if its mass is 56.0 grams.
Solution
Substitute the data into the equation:

3. Percent Error:    

Problem
A student measured the density of a sample of iron and obtained a value of 8.391 g/cm3. What is the percent error of the student’s determination?
Solution
Use Reference Table to obtain the density of iron (7.874 g/cm3).
Substitute the data into the equation:

 

4. Percent Composition:     % composition by mass =


Problem
Calculate the percent composition by mass of carbon in C6H12O6.
Solution
Use the Periodic Table to determine the mass of one mole of the compound (180.2 g) and the mass of the carbon in 1 mole of the compound (72.07 g).
Substitute the data into the equation:

 

5 a. Parts per Million:    

Problem
The concentration of arsenic in a certain river is 0.267 gram per 2,000 grams of river water.What is the arsenic concentration in parts per million (ppm)?
Solution
Substitute the data into the equation:

 

5 b. Molarity:    

Problem
What is the molarity of an aqueous solution of ammonia (gram-formula mass = 17.03 g/mol) if 30.8 grams of solute are dissolved in 2.50 liters of solution?
Solution
Use the gram-formula mass to convert the mass to moles.
Substitute the data into the equation:

 

6. Combined Gas Law:    

Problem
A 50.0-milliliter sample, initially at STP, has its pressure changed to 0.85 atmosphere and its temperature changed to 330 K.What is the new volume of the gas?
Solution
Use Reference Table A to obtain the STP values for temperature (273 K) and pressure (1 atm).
Rearrange the equation in order to solve for V2:  

Substitute the data into the equation:

 

7. Titration (Acid–Base):     MAVA = MBVB
Problem
How many milliliters of 0.25-molar NaOH are needed to neutralize 75 milliliters of 0.17-molar HCl?
Solution
Rearrange the equation to solve for VB:    
.
Substitute the data into the equation:

 

8 a. Heat Transferred:     q = mCΔT
Problem
How much heat is released by 120 grams of liquid water as it cools from 340 K to 290 K?
Solution
Use Reference Table B to obtain the specific heat capacity of liquid water (4.2 J/g·K).
Substitute the data into the equation:

(The minus sign indicates that heat is released.)
 

8 b. Heat of Fusion:    q = mHf
Problem
How much heat is absorbed by 200.0 grams of ice as it melts at 273 K?
Solution
Use Reference Table B to obtain the value for the heat of fusion of ice (333.6 J/g).
Substitute the data into the equation:

8 c. Heat of Vaporization:    

Problem
How many grams of steam at 100°C will be condensed to liquid water if 4,000. joules of heat are released?
Solution
Use Reference Table B to obtain the value for the heat of vaporization of water (2,259 J/g).
Rearrange the equation to solve for mass:

Substitute the data into the equation:

 

9. Temperature:    K = °C + 273
Problem
What is the Kelvin temperature equivalent of 34°C?
Solution
Substitute the data into the equation:
K = 34°C + 273 = 307 K
 

10 a. Radioactive Decay:     fraction remaining =

10 b. Radioactive Decay:     number of half-life periods =

Problem
What fraction of a sample of 42K will remain unchanged after 37.2 hours of decay?
Solution

Use Reference Table N to obtain the half-life of 42K (12.4 h).

Use equation 10b to determine the number of half-life periods:

Substitute the number of half-life periods into equation 10 a:
fraction remaining =

That is, one-eighth of the original sample of 42K remains unchanged after 37.2 hours of decay.