Problems (60 points correct (60 points total) Be sure to show your work and give
ID: 1029840 • Letter: P
Question
Problems (60 points correct (60 points total) Be sure to show your work and give answers to number of significant figures and with proper units. Round atomie weights and molar masses to the tenths place. 1. (10 points) Consider a 9.58 g sample of the iodine trioxide 1Os a) How many moles of compound are present in the sample? b) How many grams of iodine are present in the sample? c) How many moles of oxygen atoms are present in the samplet d) How many atoms of iodine are present in the sample? e) How many atoms of oxygen are present in the samplie?Explanation / Answer
a)molar mass of IO3 = 174.903 g/mol
no. of mole = gm of compound / molar mass
no. of mole of IO3 = 9.58 / 174.903 = 0.05477 mole
in 9.58 gm IO3 contain 0.05477 mole of IO3
b) molar mass of IO3 = 174.903 g/mol
molar mass of I = 126.90447 g/mol
in 1 mole of IO3 contain 1 mole of I that mean in 174.903 gm IO3 contain 126.90447 gm of I then 9.58 gm of IO3 contain I = 126.90447 X 9.58 / 174.90447 = 6.95 gm
in 9.58 gm IO3 present 6.95 gm iodine
c)
molar mass of IO3 = 174.903 g/mol
molar mass of O3 = 48 g/mol
in 1 mole of IO3 contain 1 mole of O3 that mean in 174.903 gm IO3 contain 48 gm of I then 9.58 gm of IO3 contain O3 = 48 X 9.58 / 174.90447 = 2.63 gm
in 9.58 gm IO3 present 2.63 gm oxygen
d) as calculated above in 9.58 gm sample of IO3 contain 6.95 gm of Iodine
molar mass of I = 126.90447 g/mol
no. of mole = gm of compound / molar mass
no. of mole of iodine = 6.95 / 126.90447 = 0.0547656 mole
according to avogadro's law 1 mole contain 6.022 X 1023 atom
no. of atom = no. of mole X 6.022 X 1023
no. of atom of iodine = 0.0547656 X 6.022 X 1023 = 3.298 X 1022 atom
no. of atom of iodine = 3.298 X 1022 atom
e)
as calculated above in 9.58 gm sample of IO3 contain 2.63 gm of oxygen
molar mass of O = 16 g/mol
no. of mole = gm of compound / molar mass
no. of mole of oxygen = 2.63 / 16 = 0.164375 mole
according to avogadro's law 1 mole contain 6.022 X 1023 atom
no. of atom = no. of mole X 6.022 X 1023
no. of atom of oxygen = 0.164375 X 6.022 X 1023 = 9.899 X 1022 atom
no. of atom of oxygen = 9.899 X 1022 atom
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