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Question
15 déterminez la formule chimique des composés formés à partir des substances suivantes. pour ce faire, remplissez le tableau suivant. substances ion positif ion négatif formule chimique du composé formé a) du lithium avec du sulfite. b) du phosphate avec du calcium. c) du magnésium avec de l’hydrogène. d) du chlore avec de l’ammonium. e) de l’aluminium avec de l’oxygène.
To determine the chemical formula of the compound formed, we use the cross - multiplication method (also known as the criss - cross method) for the positive and negative ions. The principle is that the magnitude of the charge of the positive ion becomes the subscript of the negative ion, and the magnitude of the charge of the negative ion becomes the subscript of the positive ion. Then we simplify the subscripts to the simplest integer ratio.
Part (a) Lithium and Sulfite
- Step 1: Identify the ions
The positive ion (cation) is $\ce{Li+}$ (lithium ion) and the negative ion (anion) is $\ce{SO_3^{2 - }}$ (sulfite ion).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Li+}$ is $+ 1$ and the charge of $\ce{SO_3^{2 - }}$ is $- 2$. We cross - multiply the magnitudes of the charges. So the subscript of $\ce{Li}$ will be $2$ (the magnitude of the charge of $\ce{SO_3^{2 - }}$) and the subscript of $\ce{SO_3}$ will be $1$ (the magnitude of the charge of $\ce{Li+}$).
The chemical formula is $\ce{Li_2SO_3}$.
Part (b) Calcium and Phosphate
- Step 1: Identify the ions
The positive ion is $\ce{Ca^{2+}}$ (calcium ion) and the negative ion is $\ce{PO_4^{3 - }}$ (phosphate ion).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{PO_4^{3 - }}$ is $- 3$. Cross - multiplying the magnitudes of the charges, the subscript of $\ce{Ca}$ is $3$ (the magnitude of the charge of $\ce{PO_4^{3 - }}$) and the subscript of $\ce{PO_4}$ is $2$ (the magnitude of the charge of $\ce{Ca^{2+}}$).
The chemical formula is $\ce{Ca_3(PO_4)_2}$.
Part (c) Magnesium and Hydrogen
- Step 1: Identify the ions
The positive ion is $\ce{H+}$ (hydrogen ion) and the negative ion is $\ce{Mg^{2+}}$ (magnesium ion). But we should note that in the formation of a compound, hydrogen can act as a cation ($\ce{H+}$) and magnesium as a cation. Wait, actually, when magnesium reacts with hydrogen, magnesium will lose electrons to form $\ce{Mg^{2+}}$ and hydrogen will gain electrons to form $\ce{H-}$ (hydride ion). Maybe there was a mistake in the ion labeling in the original problem. Assuming the correct ions: positive ion $\ce{Mg^{2+}}$, negative ion $\ce{H-}$.
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Mg^{2+}}$ is $+ 2$ and the charge of $\ce{H-}$ is $- 1$. Cross - multiplying, the subscript of $\ce{Mg}$ is $1$ (the magnitude of the charge of $\ce{H-}$) and the subscript of $\ce{H}$ is $2$ (the magnitude of the charge of $\ce{Mg^{2+}}$).
The chemical formula is $\ce{MgH_2}$.
Part (d) Ammonium and Chlorine
- Step 1: Identify the ions
The positive ion is $\ce{NH_4^{+}}$ (ammonium ion) and the negative ion is $\ce{Cl-}$ (chloride ion, since chlorine gains an electron to form $\ce{Cl-}$).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{NH_4^{+}}$ is $+ 1$ and the charge of $\ce{Cl-}$ is $- 1$. Cross - multiplying, the subscript of $\ce{NH_4}$ is $1$ (the magnitude of the charge of $\ce{Cl-}$) and the subscript of $\ce{Cl}$ is $1$ (the magnitude of the charge of $\ce{NH_4^{+}}$).
The chemical formula is $\ce{NH_4Cl}$.
Part (e) Aluminum and Oxygen
- Step 1: Identify the ions
The positive ion is $\ce{Al^{3+}}$ (aluminum ion) and the negative ion is $\ce{O^{2 - }}$ (oxide ion).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Al^{3+}}$ is $+ 3$ and the charge of $\ce{O^{2 - }}$ is $- 2$. Cross - multiplying the magnitudes of the charges, the subscript of $\ce{Al}$ is $2$ (the magnitude of the…
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To determine the chemical formula of the compound formed, we use the cross - multiplication method (also known as the criss - cross method) for the positive and negative ions. The principle is that the magnitude of the charge of the positive ion becomes the subscript of the negative ion, and the magnitude of the charge of the negative ion becomes the subscript of the positive ion. Then we simplify the subscripts to the simplest integer ratio.
Part (a) Lithium and Sulfite
- Step 1: Identify the ions
The positive ion (cation) is $\ce{Li+}$ (lithium ion) and the negative ion (anion) is $\ce{SO_3^{2 - }}$ (sulfite ion).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Li+}$ is $+ 1$ and the charge of $\ce{SO_3^{2 - }}$ is $- 2$. We cross - multiply the magnitudes of the charges. So the subscript of $\ce{Li}$ will be $2$ (the magnitude of the charge of $\ce{SO_3^{2 - }}$) and the subscript of $\ce{SO_3}$ will be $1$ (the magnitude of the charge of $\ce{Li+}$).
The chemical formula is $\ce{Li_2SO_3}$.
Part (b) Calcium and Phosphate
- Step 1: Identify the ions
The positive ion is $\ce{Ca^{2+}}$ (calcium ion) and the negative ion is $\ce{PO_4^{3 - }}$ (phosphate ion).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Ca^{2+}}$ is $+ 2$ and the charge of $\ce{PO_4^{3 - }}$ is $- 3$. Cross - multiplying the magnitudes of the charges, the subscript of $\ce{Ca}$ is $3$ (the magnitude of the charge of $\ce{PO_4^{3 - }}$) and the subscript of $\ce{PO_4}$ is $2$ (the magnitude of the charge of $\ce{Ca^{2+}}$).
The chemical formula is $\ce{Ca_3(PO_4)_2}$.
Part (c) Magnesium and Hydrogen
- Step 1: Identify the ions
The positive ion is $\ce{H+}$ (hydrogen ion) and the negative ion is $\ce{Mg^{2+}}$ (magnesium ion). But we should note that in the formation of a compound, hydrogen can act as a cation ($\ce{H+}$) and magnesium as a cation. Wait, actually, when magnesium reacts with hydrogen, magnesium will lose electrons to form $\ce{Mg^{2+}}$ and hydrogen will gain electrons to form $\ce{H-}$ (hydride ion). Maybe there was a mistake in the ion labeling in the original problem. Assuming the correct ions: positive ion $\ce{Mg^{2+}}$, negative ion $\ce{H-}$.
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Mg^{2+}}$ is $+ 2$ and the charge of $\ce{H-}$ is $- 1$. Cross - multiplying, the subscript of $\ce{Mg}$ is $1$ (the magnitude of the charge of $\ce{H-}$) and the subscript of $\ce{H}$ is $2$ (the magnitude of the charge of $\ce{Mg^{2+}}$).
The chemical formula is $\ce{MgH_2}$.
Part (d) Ammonium and Chlorine
- Step 1: Identify the ions
The positive ion is $\ce{NH_4^{+}}$ (ammonium ion) and the negative ion is $\ce{Cl-}$ (chloride ion, since chlorine gains an electron to form $\ce{Cl-}$).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{NH_4^{+}}$ is $+ 1$ and the charge of $\ce{Cl-}$ is $- 1$. Cross - multiplying, the subscript of $\ce{NH_4}$ is $1$ (the magnitude of the charge of $\ce{Cl-}$) and the subscript of $\ce{Cl}$ is $1$ (the magnitude of the charge of $\ce{NH_4^{+}}$).
The chemical formula is $\ce{NH_4Cl}$.
Part (e) Aluminum and Oxygen
- Step 1: Identify the ions
The positive ion is $\ce{Al^{3+}}$ (aluminum ion) and the negative ion is $\ce{O^{2 - }}$ (oxide ion).
- Step 2: Apply the cross - multiplication method
The charge of $\ce{Al^{3+}}$ is $+ 3$ and the charge of $\ce{O^{2 - }}$ is $- 2$. Cross - multiplying the magnitudes of the charges, the subscript of $\ce{Al}$ is $2$ (the magnitude of the charge of $\ce{O^{2 - }}$) and the subscript of $\ce{O}$ is $3$ (the magnitude of the charge of $\ce{Al^{3+}}$).
The chemical formula is $\ce{Al_2O_3}$.
Filling the table:
| Substances | Ion positif | Ion négatif | Formule chimique du composé formé |
|---|---|---|---|
| b) Du phosphate avec du calcium. | $\ce{Ca^{2+}}$ | $\ce{PO_4^{3 - }}$ | $\ce{Ca_3(PO_4)_2}$ |
| c) Du magnésium avec de l’hydrogène. | $\ce{Mg^{2+}}$ | $\ce{H-}$ (corrected) | $\ce{MgH_2}$ |
| d) Du chlore avec de l’ammonium. | $\ce{NH_4^{+}}$ | $\ce{Cl-}$ | $\ce{NH_4Cl}$ |
| e) De l’aluminium avec de l’oxygène. | $\ce{Al^{3+}}$ | $\ce{O^{2 - }}$ | $\ce{Al_2O_3}$ |