How many grams are there in 3.3 x 10²³ molecules of N₂I₆?
To calculate the mass of 3.3 x 10²³ molecules of N₂I₆, we first need to find the molar mass of N₂I₆.
The molar mass of N₂I₆ can be calculated by adding the individual molar masses of nitrogen (N) and iodine (I) and then multiplying by the respective subscripts in the formula.
Molar mass of N = 14.007 g/mol
Molar mass of I = 126.904 g/mol
Molar mass of N₂I₆ = (2 * molar mass of N) + (6 * molar mass of I)
Molar mass of N₂I₆ = (2 * 14.007 g/mol) + (6 * 126.904 g/mol)
Molar mass of N₂I₆ = 28.014 g/mol + 761.424 g/mol
Molar mass of N₂I₆ = 789.438 g/mol
Now, we can calculate the mass of 3.3 x 10²³ molecules of N₂I₆ using Avogadro's number.
1 mol of N₂I₆ = 789.438 g
3.3 x 10²³ molecules of N₂I₆ = (3.3 x 10²³) / 6.022 x 10²³ mol
3.3 x 10²³ molecules of N₂I₆ = 0.547 mol
Mass of 3.3 x 10²³ molecules of N₂I₆ = 0.547 mol * 789.438 g/mol
Mass of 3.3 x 10²³ molecules of N₂I₆ = 432.106986 g
Therefore, there are 432.11 grams in 3.3 x 10²³ molecules of N₂I₆.