$解：设石灰石样品中碳酸钙的质量为x。$
$\ \text {CaCO}_3+2\ \text {HCl}\xlongequal[\ \ \ \ \ \ ]{\ \ \ \ \ \ }\ \text {CaCl}_2+\ \text H_2\ \text {O}+\ \text {CO}_2↑$
$\mathrm{ \frac {100 }{ x}=\frac { 44 }{ 3.3g } } ，解得 x = 7.5\ \mathrm {g}$
$则石灰石样品中碳酸钙的质量分数为\mathrm{\frac{7.5 g}{10 g}×100％=75 ％}$
$答：石灰石样品中碳酸钙的质量分数为75％。$
