Question #6f50c

You need to use titanium's atomic mass to figure out the mass and abundance of its fifth isotope.

So, start with what you know

• #""^46"Ti" -># #"45.953 u"# and #8.0%#
• #""^47"Ti" -># #"46.592 u"# and #7.3%#
• #""^48"Ti" -># #"47.948 u"# and #73.8%#
• #""^49"Ti" -># #"48.948 u"# and #5.5%#

Now take a look at a periodic table. Titanium's molar mass is known to be #"47.867 g/mol"#.

If you take into account the fact that one unified atomic mass unit, or #u# (amu is a very old term that is sometimes used interchangeably with #u#, although it was not defined the same), is equivalent to #"1 g/mol"#, you can say that titanium has a relative atomic mass of

#47.867color(red)(cancel(color(black)("g/mol"))) * "1 u"/(1color(red)(cancel(color(black)("g/mol")))) = "47.867 u"#

The abundances of the five stable isotopes must add up #100%#, which means that the percent abundance of the fifth isotope will be

#%""^50"Ti" = 100% - (8.0 + 7.3 + 73.8 + 5.5)% = 5.4%#

The relative atomic mass of titanium is the sum of each isotope's atomic mass multiplied by its abundance

#"relative atomic mass" = sum_i("isotope"""_i * "abundance"""_i)#

This means that you have - I'll use fractional abundances, which are simply percent abundances divided by #100#

#45.953 * 0.08 + 46.952 * 0.073 + 47.948 * 0.738 + 48.948 * 0.055 + color(blue)(x) * 0.054 = 47.867#

This means that #color(blue)(x)#, which represents the atomic mass of the fifth isotope, will be

#color(blue)(x) * 0.054 = 47.867 - 45.1552#

#color(blue)(x) = 2.71178/0.054 = "50.218 u"#

SIDE NOTE THe actual atomic mass of titanium's fifth stable isotope, titanium-50, is #"49.945 u"#.

From what I can tell, the difference between this result and the actual value comes from the way the values given to you for the abundances have been rounded.