Respuesta :
The mass of each substance can be found through stoichiometric calculation as follows:
[tex] \frac{g}{u} = \frac{a}{C} [/tex]
where "g" is the mass of substance, "u" is the relative atomic mass, "a" is the number of atoms or molecules and "C" is Avogadro's number.
Making use of that formula, we have:
a) 6.02 x 10²³ atoms of Mg.
[tex] \frac{g}{24.3}= \frac{6.02\times 10^{23}}{6.02 \times10^{23}} \\ \boxed {g= 24.3~g}[/tex]
b) 12.4 x 10¹⁵ molecules of CH₂O.
[tex]\frac{g}{12+2 \times 1+16}= \frac{12.4\times 10^{25}}{6 \times10^{23}} \\ g= \frac{30 \times 12.4 \times 10^{25}}{6 \times10^{23}} \\ \boxed {g= 6.2~kg}[/tex]
c) 3.01 x 10²³ formula units of CaCl₂
[tex]\frac{g}{40+2 \times 35.5}= \frac{3.01 \times 10^{25}}{6 \times10^{23}} \\ g= \frac{111 \times 3.01 \times 10^{25}}{6 \times10^{23}} \\ \boxed {g= 5.5685~kg}[/tex]
[tex] \frac{g}{u} = \frac{a}{C} [/tex]
where "g" is the mass of substance, "u" is the relative atomic mass, "a" is the number of atoms or molecules and "C" is Avogadro's number.
Making use of that formula, we have:
a) 6.02 x 10²³ atoms of Mg.
[tex] \frac{g}{24.3}= \frac{6.02\times 10^{23}}{6.02 \times10^{23}} \\ \boxed {g= 24.3~g}[/tex]
b) 12.4 x 10¹⁵ molecules of CH₂O.
[tex]\frac{g}{12+2 \times 1+16}= \frac{12.4\times 10^{25}}{6 \times10^{23}} \\ g= \frac{30 \times 12.4 \times 10^{25}}{6 \times10^{23}} \\ \boxed {g= 6.2~kg}[/tex]
c) 3.01 x 10²³ formula units of CaCl₂
[tex]\frac{g}{40+2 \times 35.5}= \frac{3.01 \times 10^{25}}{6 \times10^{23}} \\ g= \frac{111 \times 3.01 \times 10^{25}}{6 \times10^{23}} \\ \boxed {g= 5.5685~kg}[/tex]
