How many bromine atoms are present in 33.8 g of ch2br2?

2.34 x 10^23 Calculate the molar mass of CH2Br2 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight Bromine = 79.904 Molar mass of CH2Br2 = 12.0107 + 2 * 1.00794 + 2 * 79.904 = 173.83458 Now divide the given mass of CH2Br2 you have by the molar mass 33.8 g / 173.83458 g/mol = 0.194438 moles Since each molecule of CH2Br2 has 2 bromine atoms, multiply by 2 0.194438 * 2 = 0.388876 moles Finally, multiply by avogadro's number 0.388876 * 6.0221409 x 10^23 = 2.341866 x 10^23 Since you only have 3 significant figures, round the result to 3 significant figures. 2.341866 x 10^23 = 2.34 x 10^23