Answer
$C_{Fe} $= 0.9887 wt %
$C_{C} $= 0.0019 wt%
$C_{Cr} $ = 0.0094 wt%
Work Step by Step
Given:
alloy that contains 105 kg of Fe, 0.2 kg of C, and 1.0 kg of Cr
Required:
composition by weight
Solution:
Using the modified form of Equation 4.3:
For Fe:
$C_{Fe} = \frac{m_{Fe}}{m_{Fe} + m_{C} +m_{Cr}} \times 100 = \frac{105 kg}{105 kg + 0.2 kg +1.0 kg} \times 100 $= 98.87 wt%
For C,
$C_{C} = \frac{m_{C}}{m_{Fe} + m_{C} +m_{Cr}} \times 100 = \frac{0.2 kg}{105 kg + 0.2 kg +1.0 kg} \times 100 $= 0.19 wt%
For Cr,
$C_{Cr} = \frac{m_{Cr}}{m_{Fe} + m_{C} +m_{Cr}} \times 100 = \frac{1.0 kg}{105 kg + 0.2 kg +1.0 kg} \times 100 $= 0.94 wt%
