解:
设$50\ \mathrm {kg }$碳酸氢铵化肥中至少含有纯净的$NH_{4}HCO_{3}$的质量为$x。$
$NH_{4}HCO_{3}$中氮元素的质量分数$=\frac {14}{14+1×5+12+16×3}×100\%=\frac {14}{79}×100\%$
已知化肥含氮量$ 16.0\%,$则$x×\frac {14}{79}×100\%=50\ \mathrm {kg}×16.0\%$
$x=\frac {50\ \mathrm {kg}×16.0\%×79}{14}≈45.1\ \mathrm {kg}$
答:$50\ \mathrm {kg }$碳酸氢铵化肥中至少含有纯净的$NH_{4}HCO_{3}$的质量约为$45.1\ \mathrm {kg}。$