First of all we need to look at the decomposition equation of limestone. We see that calcium carbonate (calcite) decomposes in a 1:1 ratio to CaO and CO2 - i.e. for every mole of CaCO3 you decompose you will produce 1 mole of CaO and 1 mole of CO2.
Knowing this we can use the number of moles of CO2 produced as evidence for the number of moles CaCO3 in the original sample (because these values will be the same). If we can assume that CO2 is an ideal gas (usually the case), we can say:
PV = nRT (this is the ideal gas equation)
P = pressure, V = volume, n = Number of Moles (what we will solve for), R = ideal gas constant, T = temperature. At STP we know that P = 100,000 Pa and T = 273.15 K, and we are given volume and R is a constant, therefore we can solve for n.
Firstly however, if we are going to use P in pascals, T in k, R in J/mole*k, and n in moles (units), then we need to have the volume in cubic meters. Therefore:
V = 284 mL = 284 cm cubed
284 cm cubed * 10 to the negative 6 meters cubed per cm cubed = 0.000284 meters cubed.
Now we can rearrange the gas law to solve for n,
n = PV / RT = (100,000*0.000284) / (8.314510*273.15) = 0.0125 moles
note: do not worry about where the 8.314510 came from it is just a constant.
Because we know that the reaction produced 0.0125 moles of CO2, we know that there must have been this amount of calcite in the original sample. Therefore if we multiply this number by the molecular weight of calcite we will get the number of grams of calcite in the original sample - i.e.:
0.0125 moles * 100.09 g / mole = 1.25 g
Therefore we had 1.25 g CaCO3 in our original sample. Therefore the mass percentage is given by:
MP% = mass CaCO3 / mass sample * 100%
= 1.25 g / 1.506 g * 100% = 83.11% (by mass)