# Homework 21

Q1.64

Question:

Isotopes are atoms that have the same number of protons but differ in the number of neutrons. Write the chemical symbols for each of the following isotopes:

a) the chlorine isotope with 18 neutrons

b) the chlorine isotope with 20 neutrons

c) the calcium isotope with 20 neutrons

d) the neon isotope with 11 neutrons

Strategy:

Step 1: Determine the given element's atomic number from the periodic table.

Step 2: Add the number of neutrons given in the question to the atomic number to calculate the mass number of the isotope.

Step 3: The mass number will be represented in the upper left corner of the chemical symbol, while the atomic number will be represented in the lower left corner.

• A chemical symbol is written as $_{Z}^{A}\textrm{X}$ . The A stands for the number of protons and neutrons, also known as the mass number. The Z stands for the number of protons, also known as the atomic number.
• Subtracting Z from A will give you the number of neutrons.
• Example: $_{19}^{39}\textrm{K}$ --> 39-19= 20 neutrons
• Therefore, adding the given number of neutrons to the atomic number, Z, gives you the mass number.

Solution:

a) $_{17}^{35}\textrm{Cl}$

b) $_{17}^{37}\textrm{Cl}$

c) $_{20}^{40}\textrm{Ca}$

d) $_{10}^{21}\textrm{Ne}$

Q5.80

Question:

For each substance, convert the given molecules to mass in grams:

a) 3.2 x 1024 Cl2 molecules

b) 8.25 x 1018 CH2O molecules

c) 1 carbon dioxide molecule

Strategy:

Step 1: Convert the given molecules to moles by dividing by Avogadro's number.

Step 2: Multiply the number of moles by the Molar Mass of the substance to determine the grams.

Step 3: Through correct Dimensional Analysis, the molecules and moles will cancel to leave grams.

$\frac{molecules}{Avogadro's}\rightarrow moles\rightarrow moles \cdot MM\rightarrow grams$

• Avogadro's number: 6.022 x 1023 mol of a substance
• Molar mass (MM): the weight of a substance in the units g/mol, found by adding the atomic masses of each element.
• Example: MM of H2O --> H(1.008 x 2) + O(15.999)= 18 g/mol

Solution:

a) 3.8 x 1051 grams

b) 4.11 x 1045 grams

c) 7.31 x 1025 grams