Solution to standard addition challenge

where Ri is the ratio of the signal of the analyte to that of the internal standard. Note that in this way the method becomes independent of temperature (density), and that storage or loss of the samples will not affect the result. The “trick” in calculating the mass fraction from experiment 2 is to note that because the internal standard is already present in the original sample the masses of the sample solution (mx,i) and the internal standard solution (my,i) are equal, because both form the same solution. Thus: yi 1⁄4 Ri my;i mx;i 1⁄4 Ri mx;i mx;i 1⁄4 Ri: