# All of the lanthanide metals react with HCl to form compounds having the formula MCl_2, MCl_3,or...

## Question:

All of the lanthanide metals react with {eq}\rm HCl{/eq} to form compounds having the formula {eq}\rm MCl_2,\ MCl_3,{/eq} or {eq}\rm MCl_4{/eq} (where {eq}\rm M{/eq} represents the metallic element). Each metal forms a single compound. A chemist has {eq}0.250\ g{/eq} sample of a transition metal and wishes to identify the metal. She reacts the metal with excess {eq}\rm HCl{/eq} and obtains {eq}0.427\ g{/eq} of product. Based on this information, identify the metal and write the chemical formula of the product.

## Molecular Weight:

Molecular weight is the bridge connecting the molar amount and the mass of a substance. The molecular weight of a substance is the total sum of all the elements in its chemical structure.

## Answer and Explanation:

In this reaction:

{eq}2M + xCl_2 \rightarrow 2MCl_x \\ x = 2 \ or \ 3 \ or \ 4{/eq}

The mass difference obtained in the upper reaction is the mass of {eq}Cl_2{/eq}, thus the mass of {eq}Cl_2{/eq} is:

{eq}0.427 - 0.250 = 0.177 \ g{/eq}

The molar amount of {eq}Cl{/eq} element is:

{eq}0.177 / 35.4 = 0.005 \ mol{/eq}

If we assume {eq}x = 2{/eq}, then the molar amount of {eq}M{/eq} is {eq}0.005 / 2 = 0.0025 \ mol{/eq}. Thus the atomic weight of {eq}M{/eq} is {eq}0.250 / 0.0025 = 100 \ g/mol{/eq}.

No lanthanide metals have an atomic mass of 100 g/mol.

If {eq}x = 3{/eq}, then the molar amount of {eq}M{/eq} is {eq}0.005 / 3 = 0.00167 \ mol{/eq}. Thus the atomic weight of {eq}M{/eq} is {eq}0.250 / 0.00167 = 150 \ g/mol{/eq}.

Samarium (Sm) has an atomic mass of 150 g/mo.

If we assume {eq}x = 4{/eq}, then the molar amount of {eq}M{/eq} is {eq}0.005 / 4 = 0.00125 \ mol{/eq}. Thus, the atomic weight of {eq}M{/eq} is {eq}0.250 / 0.00125 = 200 \ g/mol{/eq}.

No lanthanide metals have an atomic mass of 200 g/mol.

Thus, **this metal is samarium (Sm) and the chemical formula of this product is** {eq}\mathbf{SmCl_3}{/eq}.

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