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An analyte gives \( 4500 \) area units and an internal standard \( 3000 \). Their ratio is

  • A.7500
  • B.1500
  • C.1.5
  • D.0.67

Answer: C

Dividing analyte by standard:

\[ \frac{4500}{3000} = 1.5 \]

giving 1.5, which is compared against a calibration of the same ratio to obtain the concentration. Inverting gives 0.67, and the sum and difference are not used. That the ratio rather than either area is carried forward is exactly what makes the injection volume irrelevant.

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