Download CHEM1405: Answers to Problem Sheet 1 - Elements, Compounds, Molar Masses, Solutions and more Exercises Chemistry in PDF only on Docsity!
CHEM1405 Answers to Problem Sheet 1
liquid mercury element ice molecular compound
neon gas element liquid nitrogen element
milk (^) mixture copper pipe (^) element
blood mixture air mixture
gaseous CO 2 molecular compound gaseous oxygen element
solid sodium element brass mixture
- 23490 Th : the number of neutrons is 234 – 90 = 144.
- O2–, F–^ and Ne have exactly 10 electrons.
O2–^ Atomic number 8 -2 anion has 10e– He Atomic number 2 2e– Ar Atomic number 18 18e– F–^ Atomic number 9 -1 anion has 10e– Sr Atomic number 38 38e– S2–^ Atomic number 16 -2 anion has 18e– Cl–^ Atomic number 17 -1 anion has 18e– O Atomic number 8 8e– F Atomic number 9 9e– Ne Atomic number 10 10e–
- (c) chromium, manganese, iron, cobalt, nickel
- (d) fluorine, chlorine, bromine, iodine
- Molecular mass of CH 3 NH 2 :
12.01 (C) + 3 ×××× 1.01 (H) + 14.01 (N) + 2 ×××× 1.01 (H) = 31.06 g mol-
Number of moles in 1 g:
number of moles = mass / molar mass = 1 / 31.06 = 0.03 mol
Note that the question asks for the number of moles in 1 g. Since this mass is given to only one significant figure, so is the answer.
- Molar mass of CuSO 4 ⋅⋅⋅⋅5H 2 O:
63.55 (Cu) + 32.07 (S) + 4 ×××× 16.00 (O) + 5 ×××× [2 ×××× 1.01 (H) + 16.00 (O)]
= 249.72 g mol-
Number of moles in 24.9 g of CuSO 4 .5H 2 O:
number of moles = mass / molar mass = 24.9 / 249.72 = 0.100 mol.
1 mol of CuSO 4 ⋅⋅⋅⋅5H 2 O contains 1 mol of copper so,
number of moles of copper = 0.100 mol
Note that the question gave the mass as 24.9 g – three significant figures. The answer reflects this. The trailing zeros in 0.100 imply that the number is known to three significant figures.
- The relative atomic mass of silicon is the weighted average of the masses of its isotopes:
1
(25.80) (1.36) (0.926) 28.09 g mol−−−−
××××^ ++++^ ××××^ ++++^ ××××
The numbers in brackets are given to four, three and three significant figures respectively since this is the precision of the relative abundances in the question. When these are added, the answer is precise to the second decimal place as this is where each term is known precisely.
- (a) As density is given by
mass (m) density ( ρ ) volume (V)
The mass of 1.00 L ( = 1000 mL) of water is:
m ==== ρ ×××× V ==== 0.997 ×××× 1000 ====997 g
The molar mass of H 2 O is:
2 ×××× 1.01 (H) + 16.00 (O)=18.02 g mol-
Hence, the number of moles in 997 g is:
mass (m) 997 number of moles (n) 55.3mol molar mass (M) 18.
