Question

1. calculate the average atomic mass of gold with the 50% being gold-197 and 50% being gold-198.
2. calculate the average atomic mass of lithium, which occurs as two isotopes that have the following atomic masses and abundances in nature: 6.017 g, 7.30% and 7.018 g, 92.70%.
3. hydrogen is 99% ¹h, 0.8% ²h, and 0.2% ³h. calculate its average atomic mass.

Explanation:

3.

Step1: Write the formula for average atomic mass

The formula is $Average\ atomic\ mass=\sum_{i}(Mass_{i}\times Abundance_{i})$. Here, $i$ represents each isotope. For gold, we have two isotopes with equal abundance.

Step2: Substitute values

Gold - 197 has a mass of 197 and abundance of 0.5, gold - 198 has a mass of 198 and abundance of 0.5. So, $Average\ atomic\ mass=(197\times0.5)+(198\times0.5)$.

Step3: Calculate the result

$197\times0.5 = 98.5$, $198\times0.5=99$, and $98.5 + 99=197.5$.

Step1: Apply the average - atomic - mass formula

$Average\ atomic\ mass=(Mass_1\times Abundance_1)+(Mass_2\times Abundance_2)$.

Step2: Substitute the given values

$Mass_1 = 6.017\ g$, $Abundance_1=0.0730$, $Mass_2 = 7.018\ g$, $Abundance_2 = 0.9270$. So, $Average\ atomic\ mass=(6.017\times0.0730)+(7.018\times0.9270)$.

Step3: Calculate each product

$6.017\times0.0730 = 0.439241$, $7.018\times0.9270=6.495686$.

Step4: Sum the products

$0.439241+6.495686 = 6.934927\approx6.935\ g$.

Step1: Use the average - atomic - mass formula

$Average\ atomic\ mass=(Mass_1\times Abundance_1)+(Mass_2\times Abundance_2)+(Mass_3\times Abundance_3)$. The mass of $^1H$ is 1 g/mol, $^2H$ is 2 g/mol, and $^3H$ is 3 g/mol.

Step2: Substitute the values

$Abundance_1 = 0.99$, $Mass_1 = 1$; $Abundance_2=0.008$, $Mass_2 = 2$; $Abundance_3 = 0.002$, $Mass_3 = 3$. So, $Average\ atomic\ mass=(1\times0.99)+(2\times0.008)+(3\times0.002)$.

Step3: Calculate each product

$1\times0.99 = 0.99$, $2\times0.008 = 0.016$, $3\times0.002=0.006$.

Step4: Sum the products

$0.99+0.016 + 0.006=1.012\ g/mol$.

Answer:

197.5

4.