# Question #6f50c

##### 1 Answer

#### Explanation:

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

If you take into account the fact that one **unified atomic mass unit**, or *amu* is a very old term that is sometimes used interchangeably with

#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** *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

#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) * 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.*