解:共产生$CO_{2}$的质量为$2.2\ g + 2.2\ g + 1.1\ g = 5.5\ g。$
设$15\ g$该样品中$CaCO_{3}$的质量为$x。$
$CaCO_{3} + 2HCl = CaCl_{2} + H_{2}O + CO_{2}\uparrow$
$100$ $44$
$x$ $5.5\ g$
$\frac{100}{44}=\frac{x}{5.5\ g}$,解得 $x = 12.5\ g$
该样品中$CaCO_{3}$的质量分数为$\frac{12.5\ g}{15\ g}×100\%≈83.3\%。$
答:该样品中碳酸钙的质量分数约为$83.3\%。$