First, we need to determine the moles of carbon in the sample. We can do this using the mass of carbon and its atomic mass.moles of C = mass of C / atomic mass of Cmoles of C = 3.06 g / 12.01 g/mol = 0.255 molSince there is one carbon atom in both calcium carbonate CaCO3 and magnesium carbonate MgCO3 , the moles of carbon will be equal to the total moles of CaCO3 and MgCO3 in the sample.Now, let's denote the moles of CaCO3 as x and the moles of MgCO3 as y. We can write the following equation:x + y = 0.255 molNext, we need to find the mass of CaCO3 and MgCO3 in the sample. We can do this using their molar masses:mass of CaCO3 = x * 40.08 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = x * 100.09 g/molmass of MgCO3 = y * 24.31 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = y * 84.32 g/molThe total mass of the sample is 5 grams, so we can write another equation:mass of CaCO3 + mass of MgCO3 = 5 g100.09x + 84.32y = 5Now we have a system of two equations with two variables:x + y = 0.255100.09x + 84.32y = 5We can solve this system of equations to find the values of x and y. First, we can solve the first equation for x:x = 0.255 - yNow substitute this expression for x into the second equation:100.09 0.255 - y + 84.32y = 5Simplify and solve for y:25.523 - 100.09y + 84.32y = 5-15.77y = -20.523y = 1.301 molNow substitute the value of y back into the equation for x:x = 0.255 - 1.301x = -1.046 molSince we cannot have a negative number of moles, there must be an error in the given information. However, assuming the given information is correct, we can proceed with the calculation.Now we can find the mass of MgCO3 in the sample:mass of MgCO3 = 1.301 mol * 84.32 g/mol = 109.65 gSince the total mass of the sample is 5 grams, the mass of MgCO3 is greater than the total mass of the sample, which is not possible. Therefore, there must be an error in the given information, and the percentage by mass of magnesium in the limestone sample cannot be calculated using mass balance calculations with the provided data.