Question

carbohydrates are also known as saccharides. the word saccharide is a term derived from the greek word sakcharon, meaning sugar. though not all carbohydrates are sweet, many simple sugars, or monosaccharides, are sweet to the taste. the model shows the chemical structure of a monosaccharide called galactose. galactose has a 6-atom ring shape. this means it has 6 atoms arranged arranged in a hexagon, and each of those 6 atoms bonds with other atoms. in galactose, the 6-atom ring is formed by one oxygen atom and 5 carbon atoms. but like in this model, people dont always label the carbon atoms. what is the chemical formula of galactose? c h o

Explanation:

Step1: Count Carbon Atoms

Looking at the structure of galactose, the ring has 5 carbon atoms (the unlabeled ones in the ring) plus the one in $\ce{CH2OH}$, so total C atoms: $5 + 1 = 6$.

Step2: Count Hydrogen Atoms

Let's list the H atoms: In $\ce{CH2OH}$: 2 (from $\ce{CH2}$) + 1 (from $\ce{OH}$)? Wait, no, let's count each group. The $\ce{CH2OH}$ has 2 H (on C) + 1 H (on OH? No, $\ce{CH2OH}$ is $\ce{C}$ bonded to $\ce{H2}$, $\ce{O}$, and $\ce{H}$ (from OH). Wait, better to count each H: The ring has H atoms: let's see the structure: the atoms with H: the $\ce{HO}$ (1 H), the $\ce{CH2OH}$ (2 H on C, 1 H on OH? No, $\ce{CH2OH}$ is $\ce{C - H2 - O - H}$, so that's 3 H? Wait, no, let's do it properly. The formula for galactose is a monosaccharide, and monosaccharides have the formula $\ce{C6H12O6}$ (like glucose, galactose is an isomer of glucose). But let's count from the structure:

Carbon atoms: 6 (as per the ring: 5 in ring + 1 in $\ce{CH2OH}$).

Hydrogen atoms: Let's count each H:

- $\ce{HO}$: 1 H (attached to O? No, $\ce{HO}$ is $\ce{O - H}$, so 1 H.

- $\ce{CH2OH}$: $\ce{C}$ has 2 H, $\ce{OH}$ has 1 H: total 3 H.

- The $\ce{OH}$ on the ring (the one inside): 1 H (attached to O? No, $\ce{OH}$ is $\ce{O - H}$, so 1 H.

- The H atoms on the ring carbons: let's see the structure: the carbons in the ring (excluding O) have H: let's count the H atoms:

Looking at the structure:

- Top carbon (with $\ce{CH2OH}$): has 2 H (from $\ce{CH2}$) and 1 H? Wait, no, the structure is a hexagon (6 - atom ring: 5 C and 1 O). Let's label the carbons:

Ring atoms: O, and 5 C (let's call them C1 to C5), plus the C in $\ce{CH2OH}$ (C6).

C1: bonded to O (top), $\ce{HO}$ (left: O - H), and two H? Wait, no, the structure:

- C1 (leftmost in ring): bonded to O (top), $\ce{HO}$ (O - H, so 1 H), and two H? Wait, no, the structure shows:

H | HO

|

H - C - O (ring O)

|

H

Wait, maybe a better way: galactose is a monosaccharide with formula $\ce{C6H12O6}$. Let's count H:

Each $\ce{OH}$ group: 1 H. How many $\ce{OH}$ groups? Let's see the structure:

- $\ce{HO}$ (1)

- $\ce{CH2OH}$ (1)

- $\ce{OH}$ (inside ring, 1)

- $\ce{OH}$ (rightmost, 1)

- $\ce{OH}$ (bottom, 1)

Wait, no, the structure has:

- $\ce{HO}$ (1 H)

- $\ce{CH2OH}$: $\ce{C}$ with 2 H, $\ce{O}$ with 1 H (total 3 H)

- The $\ce{OH}$ on the ring (the one with H: 1 H)

- The H atoms on the ring carbons: let's count the H atoms not in $\ce{OH}$:

Looking at the structure, the carbons in the ring (excluding O) have H:

- C1 (left): H (top), H (bottom)

- C2 (with $\ce{OH}$): H (right)

- C3 (middle): H (top)

- C4 (right): H (bottom)

Wait, maybe I'm overcomplicating. Monosaccharides like glucose, fructose, galactose have the formula $\ce{C6H12O6}$. Let's check the number of O atoms:

- $\ce{HO}$: 1 O

- Ring O: 1 O

- $\ce{CH2OH}$: 1 O

- $\ce{OH}$ (inside ring): 1 O

- $\ce{OH}$ (rightmost): 1 O

- $\ce{OH}$ (bottom): 1 O

Total O: 6.

Carbon: 6 (5 in ring + 1 in $\ce{CH2OH}$).

Hydrogen: Let's calculate using the formula for carbohydrates: for a hexose (6 - C sugar), the formula is $\ce{C6H12O6}$. Let's verify with the structure:

Each C (except the one in $\ce{CH2OH}$) is bonded to H and OH. Wait, the $\ce{CH2OH}$ has $\ce{C - H2 - O - H}$, so that's 3 H. The other 5 C atoms: each has how many H? Let's see, in the ring, each C (except the one with $\ce{OH}$) has 2 H? No, let's count all H:

From $\ce{OH}$ groups: 5 $\ce{OH}$ groups? Wait, the structure has:

- $\ce{HO}$ (1 H)

- $\ce{CH2OH}$: $\ce{OH}$ (1 H)

- $\ce{OH}$ (inside ring, 1 H)

- $\ce{OH}$ (rightmost, 1 H)

- $\ce{OH}$ (bottom, 1 H)

Wait, that's 5 $\ce{OH}$ groups? No, the $\ce{CH2OH}$ is a group, so $\ce{CH2OH}$ has one $\ce{OH}$, and then there are $\ce{HO}$, $\ce{OH}$ (inside), $\ce{OH}$ (right), $\ce{OH}$ (bottom): total 5 $\ce{OH}$? No, the structure shows:

- $\ce{HO}$ (left)

- $\ce{CH2OH}$ (top)

- $\ce{OH}$ (inside the ring, on a C)

- $\ce{OH}$ (right)

- $\ce{OH}$ (bottom)

Wait, that's 5 $\ce{OH}$ groups? No, $\ce{CH2OH}$ is a group, so that's one $\ce{OH}$, and then $\ce{HO}$ (which is $\ce{OH}$), $\ce{OH}$ (inside), $\ce{OH}$ (right), $\ce{OH}$ (bottom): total 5 $\ce{OH}$? No, $\ce{HO}$ is $\ce{OH}$, so that's 5 $\ce{OH}$ groups? Wait, no, the structure has:

- $\ce{HO}$ (1 O, 1 H)

- $\ce{CH2OH}$ (1 O, 2 H on C, 1 H on O: total 3 H)

- $\ce{OH}$ (inside ring: 1 O, 1 H)

- $\ce{OH}$ (right: 1 O, 1 H)

- $\ce{OH}$ (bottom: 1 O, 1 H)

- Ring O: 1 O (no H)

So O atoms: 1 (HO) + 1 (CH2OH) + 1 (inside OH) + 1 (right OH) + 1 (bottom OH) + 1 (ring O) = 6 O.

C atoms: 5 (ring C) + 1 (CH2OH C) = 6 C.

H atoms: from HO: 1 H; from CH2OH: 2 (on C) + 1 (on O) = 3 H; from inside OH: 1 H; from right OH: 1 H; from bottom OH: 1 H; and the H atoms on the ring C (excluding the ones with OH):

Looking at the ring C:

- C1 (left, with HO): bonded to HO (O - H), ring O, and two H? Wait, the structure shows H above and below C1: 2 H.

- C2 (with inside OH): bonded to inside OH (O - H), ring O, and one H (on the right): 1 H.

- C3 (middle): bonded to ring O and two H? Wait, the structure shows H above C3: 1 H? No, maybe I'm wrong. Let's use the formula for monosaccharides: hexoses have formula $\ce{C6H12O6}$. So H should be 12.

Let's calculate H from the structure:

- $\ce{OH}$ groups: 5 $\ce{OH}$ groups? No, wait, $\ce{CH2OH}$ is a group, so that's one $\ce{OH}$, and then $\ce{HO}$ (which is $\ce{OH}$), $\ce{OH}$ (inside), $\ce{OH}$ (right), $\ce{OH}$ (bottom): total 5 $\ce{OH}$? No, $\ce{HO}$ is $\ce{OH}$, so that's 5 $\ce{OH}$ groups? Wait, no, the structure has:

- $\ce{HO}$ (1 H)

- $\ce{CH2OH}$: $\ce{OH}$ (1 H)

- $\ce{OH}$ (inside: 1 H)

- $\ce{OH}$ (right: 1 H)

- $\ce{OH}$ (bottom: 1 H)

That's 5 H from $\ce{OH}$ groups. Then the H on the C atoms (not in $\ce{OH}$):

- $\ce{CH2OH}$: 2 H (on C)

- C1 (left): 2 H (above and below)

- C2 (with inside OH): 1 H (right)

- C3 (middle): 1 H (above)

- C4 (right): 1 H (below)

Wait, let's add them:

From $\ce{OH}$: 5 H

From C atoms: 2 (CH2OH) + 2 (C1) + 1 (C2) + 1 (C3) + 1 (C4) = 7 H

Total H: 5 + 7 = 12. Perfect, that matches $\ce{C6H12O6}$.

So the chemical formula is $\ce{C6H12O6}$.

Answer:

$\ce{C6H12O6}$