Question

g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertainty (grams) of calcium in each average calcium mass of the sample? Assume that the relative uncertainties in atomic mass are small compared the uncertainty of the total mass.

Answer 1

Answer:

The value  is   $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

  The  molar mass of $CaCO_3$ is  $MW = 100.085 \ g/mol$

   The  total mass is  $m_g = 12.3 \ g$

   The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

 The percentage of calcium in calcite is mathematically represented as

          $C = \frac{40.07}{100.085} * 100$

          $C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

     $m= \frac{m_g}{5}$

     $m= \frac{12.3}{5}$

     $m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

        $m_c = 2.46 * \frac{40.04}{100}$

       $m_c = 0.985 \ g$

The  uncertainty of  mass of a single sample is mathematically represented as

      $k = \frac{\Delta g }{5}$

        $k = \frac{0.1 }{5}$

       $k = 0.02\ g$

The  uncertainty of  mass of calcium in a single sample is mathematically represent

         $G = \frac{0.02 * 40.04}{ 100}$

          $G = 0.00801 \ g$

Generally the average mass of calcium in each sample is  

          $L = 0.985 \pm 0.00801$