Question #f5f0c

2 Answers
Jul 25, 2016

Answer:

#73.43%#

Explanation:

The trick here is to realize that after the initial sample is heated, you will only be left with strontium oxide, #"SrO"#.

Heating the sample will consume all the strontium carbonate, #"SrCO"_3#, only to leave behind strontium oxide and give off carbon dioxide, #"CO"_2#.

In other words, the difference between the initial mass of the sample and the final mass of the is equal to the mass of carbon dioxide evolved by the reaction.

You can thus say that the reaction produced

#overbrace("1.850 g")^(color(blue)("initial mass")) - overbrace("1.445 g")^(color(purple)("final mass")) = "0.405 g"#

of carbon dioxide. Use the compound's molar mass to convert this to moles

#0.405 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.009202 moles CO"_2#

Now, write down the balanced chemical equation that describes this decomposition reaction

#"SrCO"_ (3(s)) stackrel(color(red)(Delta)color(white)(aa))(->) "SrO"_ ((s)) + "CO"_(2(g))# #uarr#

Notice that every mole of strontium carbonate that undergoes decomposition produces #1# mole of carbon dioxide. This means that the reaction consumed

#0.009202 color(red)(cancel(color(black)("moles CO"_2))) * "1 mole SrCO"_3/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.009202 moles SrCO"_3#

Next, use the compound's molar mass to convert the number of moles to grams

#0.009202 color(red)(cancel(color(black)("moles SrCO"_3))) * "147.63 g"/(1color(red)(cancel(color(black)("mole SrCO"_3)))) = "1.3585 g"#

Finally, you can say that the percent composition of strontium carbonate in the original sample was

#"% SrCO"_3 = (1.3585 color(red)(cancel(color(black)("g"))))/(1.850color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)(73.43%)color(white)(a/a)|)))#

The answer is rounded to four sig figs.

Jul 25, 2016

Answer:

73.41%

Explanation:

The amount of #CO_2# in the total mix is simply the difference between the first and second masses, as only the #CO_2# will leave as a gas upon decomposition of Sr#CO_3#.
We then use the molecular weights of Sr#CO_3# and #CO_2# to calculate how much Sr#CO_3# must have been present originally to produce the observed amount of #CO_2#.

Mass of #CO_2# in mixture: 1.850-1.445 = 0.405g

0.405g/44.01 g/mol #CO_2# = 0.0092 moles #CO_2#.

Because the ratio of #CO_2# to SrO in Sr#CO_3# is 1:1, the original moles of Sr#CO_3# must also be 0.0092. Converting this back into mass: 0.0092 Moles Sr#CO_3# x 147.63g/mol = 1.358g Sr#CO_3#.

The final desired mass ratio is then 1.358/1.850 = 0.7341,
or 73.41% Sr#CO_3#.