Question

please help me answer #2 a, b, c and d

measured value of the heat capacity be too high or too low? Explain. 2- A particular metallic element, M, has a heat capacity of 0.361.gº''K 1. It forms an oxide that contains 2.90 g of M per gram of oxygen. a) Using the law of Dulong and Petit (Cp x M = 25 J. mol-1.K-1), estimate the molar mass of the metal M. g/mol b) From the composition of the oxide (2.90 g M/g O) and the molar mass of oxygen (16.0 g/mol), determine the mass of M that combines with each mole of oxygen. g M/mol o c) Use your estimate of the atomic mass from (a) and the information from (b) to determine the empirical formula of the oxide (the moles of M that combine with each mole of O). d) Now that you know the composition of the oxide and its formula, what is the accurate value of the atomic mass of the metal M? What is the identity of M? g/mol

Answer 1

2.

a.

Using the Dulong and Petit law, the heat capacity Cp and molar mass M are related as

$C_p \times M = 25 \ J \cdot mol^{-1}\cdot K^{-1}$

Given that the heat capacity of the metal M with Molar mass M is $C_p = 0.36 \ J \cdot g^{-1} \cdot K^{-1}$

Hence, the molar mass M can be calculated as

$M = \frac{25 \ J \cdot mol^{-1} \cdot K^{-1}}{C_p} = \frac{25 \ J \cdot mol^{-1} \cdot K^{-1}}{0.36 \ J \cdot g^{-1}\cdot K^{-1}}= 69.44 \ g\cdot mol^{-1} \\ \Rightarrow M \approx 69 \ g\cdot mol^{-1}$

b.

It is given that the oxide of M contains 2.90 g of M per gram of O.

Atomic mass of O = 16.0 g/mol

Hence, the mass of M that combines with 1 mol of O i.e 16.0 g of O is

$\frac{2.90 \ g \ M}{1 \ g \ O} \times 16 \ g \ O = 46.4 \ g \ M$

Hence, 46.4 g of M will combine with 1 mol of O.

c.

Molar mass of M is determined to be 69 g/mol from part A.

Hence, number of moles of M in 46.4 g is

$moles = \frac{mass}{molar\ mass} = \frac{46.4 \ g}{69 \ g/mol}= 0.672 \ mol$

Hence, 0.672 mol of M combines with 1 mol of O.

Hence, the ratio of M to O can be written as

$\frac{M}{O}= \frac{0.672}{1}$

Now, we can divide numerator and denominator by 0.672 to convert them to nearest integers to get the empirical formula.

Hence,

$\frac{M}{O} = \frac{0.672/0.672}{1/0.672} = \frac{1}{1.48}\approx \frac{1}{1.5} \approx \frac{2}{3}$

Hence, the empirical formula of the oxide is $M_2O_3$.

Hence, two moles of M combines with 3 moles of O.

d)

Mass of 3 moles of O = $3 \ mol \times 16.0 \ g/mol = 48.0 \ g$

We know that there are 2.90 g of M per gram of O.

Hence, amount of M in grams in an oxide with 48.0 g of O is

$\frac{2.90 \ g \ M}{1 \ g \ O} \times 48.0 \ g \ O = 139.2 \ g \ M$

Note, that 2 moles of M combines with 3 moles of O (i.e. 48.0 g of O). Hence, 2 moles of M must weigh 139.2 g.

Hence, the correct molar mass of M can be calculated as

$M = \frac{139.2 \ g}{2 \ mol}= {\color{Red} 69.6 \ g\cdot mol^{-1}}$

The metal with atomic mass close to 69.6 is Gallium which actually does form oxides of the form $Ga_2O_3$.

Hence, the identity of metal M is Gallium (Ga).