If mass % of Oxygen in a monovalent metal carbonate is 48. Then find the number of atoms of metal present in 5mg of this metal carbonate sample?#( N_a = 6.022 * 10^23)#

2 Answers
Sep 4, 2017

Answer:

#6.0xx10^19# #"atoms M"#

Explanation:

We're asked to find the number of atoms of the metal present in #5# #"mg"# of a monovalent metal carbonate sample.

A monovalent metal carbonate includes a metal with a valence of one; i.e. a metal in group 1 (#"Li"#, #"Na"#, #"K"#, etc.). The general formula will be

#ul("M"_2"CO"_3#

We can use dimensional analysis to find the number of moles of the metal, #"M"#, present in the compound:

#5cancel("mg sample")((48cancel("mg O"))/(100cancel("mg sample")))((1cancel("g O"))/(10^3cancel("mg O")))((1cancel("mol O"))/(16.00cancel("g O")))((2color(white)(l)"mol M")/(3cancel("mol O"))) = color(red)(ul(0.0001color(white)(l)"mol M"#

Now, we can use Avogadro's number, #N_"A"# to find the number of atoms of #"M"# in the sample:

#color(red)(0.0001)cancel(color(red)("mol M"))((6.022*10^23color(white)(l)"atoms M")/(1cancel("mol M"))) = color(blue)(ulbar(|stackrel(" ")(" "6.0 xx 10^19color(white)(l)"atoms M"" ")|)#

which I'll just round off to #2# significant figures.

Therefore, there are #color(blue)(6.0xx10^19color(white)(l)"atoms"# of the metal present in the sample.

Sep 5, 2017

Answer:

#sf(6xx10^(19))#

Explanation:

The metal is monovalent so the formula can be written #sf(M_2CO_3)#.

The mass of oxygen in the sample is given by:

#sf(m_(O)=0.48xx5xx10^(-3)=2.4xx10^(-3)color(white)(x)g)#

To find the no. moles of oxygen we divide by the mass of 1 mole:

#sf(n_(O)=m_(O)/(A_(r)[O])=(2.4xx10^(-3))/(16.0)=0.15xx10^(-3))#

From the formula we can see that the no. moles of #sf(M)# must be 2/3 of this:

#sf(n_(M)=0.15xx10^(-3)xx(2)/(3)=0.0001color(white)(x)"mol")#

To find the no. of atoms of #sf(M)# we multiply by The Avogadro Constant:

#sf(n_(M)=6.022xx10^(23)xx0.0001=6.022xx10^(19))#