## Practice Atomic Calculations Answer Key

If you’re looking to review your understanding of atomic calculations, you’ve come to the right place. In this article, we provide you with the answer key to practice atomic calculations. Whether you’re a student studying chemistry or a professional in the field, these answers will help you check your work and improve your skills. Let’s dive in!

### Question 1:

Calculate the atomic mass of an element with the following isotopic abundances: 25% of isotope A with a mass of 10 amu, and 75% of isotope B with a mass of 12 amu.

**Answer:** The atomic mass is calculated by multiplying the mass of each isotope by its abundance and summing the results. In this case, the atomic mass would be (0.25 * 10 amu) + (0.75 * 12 amu) = 11 amu.

### Question 2:

Determine the number of protons, neutrons, and electrons in an atom with an atomic number of 20 and a mass number of 40.

**Answer:** The number of protons is equal to the atomic number, which is 20. The number of neutrons is calculated by subtracting the atomic number from the mass number, so 40 – 20 = 20 neutrons. The number of electrons is equal to the number of protons in a neutral atom, so there are 20 electrons.

### Question 3:

Calculate the average atomic mass of chlorine, given the following isotopic abundances and masses: 75% of isotope Cl-35 with a mass of 34.9689 amu, and 25% of isotope Cl-37 with a mass of 36.9659 amu.

**Answer:** The average atomic mass is calculated similarly to Question 1. Multiply the mass of each isotope by its abundance and sum the results. For chlorine, (0.75 * 34.9689 amu) + (0.25 * 36.9659 amu) = 35.5 amu.

### Question 4:

Determine the number of moles in 24 grams of magnesium (Mg).

**Answer:** To find the number of moles in a given mass of an element, divide the mass by the molar mass of the element. The molar mass of magnesium is approximately 24.305 g/mol. So, 24 g / 24.305 g/mol = 0.987 moles of magnesium.

### Question 5:

Calculate the percent composition of carbon in a compound with the formula C6H12O6.

**Answer:** To find the percent composition of an element in a compound, divide the mass of the element by the molar mass of the compound and multiply by 100. In this case, the molar mass of C6H12O6 is approximately 180.16 g/mol. The mass of carbon in the compound is 6 * 12 g/mol = 72 g/mol. Therefore, (72 g/mol / 180.16 g/mol) * 100 = 39.97% composition of carbon.

### Question 6:

Determine the empirical formula of a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

**Answer:** To find the empirical formula of a compound, convert the percentages to grams, divide by the molar mass of each element, and simplify the ratios. In this case, assume 100 g of the compound. This gives 40 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen. Dividing by the molar masses (12 g/mol for carbon, 1 g/mol for hydrogen, and 16 g/mol for oxygen) gives you a ratio of approximately 3:6:3. Simplifying the ratio, you get the empirical formula as CH2O.

### Question 7:

Calculate the wavelength of a photon with an energy of 200 kJ/mol.

**Answer:** To find the wavelength of a photon, you can use the equation E = hc/λ, where E is the energy, h is the Planck constant, c is the speed of light, and λ is the wavelength. Rearranging the equation to solve for λ gives you λ = hc/E. Plugging in the values (h = 6.626 x 10^-34 J s, c = 3.00 x 10^8 m/s, E = 200,000 J/mol) gives you a wavelength of approximately 4.96 x 10^-7 meters.

### Question 8:

Determine the frequency of the photon from Question 7.

**Answer:** The frequency of a photon can be calculated using the equation c = λν, where c is the speed of light, λ is the wavelength, and ν is the frequency. Rearranging the equation to solve for ν gives you ν = c/λ. Plugging in the values gives you a frequency of approximately 6.06 x 10^14 Hz.